L'Hospital's Rule application with raised exponents.












4












$begingroup$


I am having a little trouble going about:
$$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x}$$



Using $ln$ properties we can bring down the $10x$ exponent and have:$$ln y=10xlnleft(frac{14x}{14x+10}right)$$



And from here I get stuck trying to apply L'Hospital's Rule to find the limit.






calculus limits derivatives







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$endgroup$












  • $begingroup$
    You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
    $endgroup$
    – gimusi
    Dec 4 '18 at 20:35
















4












$begingroup$


I am having a little trouble going about:
$$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x}$$



Using $ln$ properties we can bring down the $10x$ exponent and have:$$ln y=10xlnleft(frac{14x}{14x+10}right)$$



And from here I get stuck trying to apply L'Hospital's Rule to find the limit.






calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
    $endgroup$
    – gimusi
    Dec 4 '18 at 20:35














4












4








4


1



$begingroup$


I am having a little trouble going about:
$$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x}$$



Using $ln$ properties we can bring down the $10x$ exponent and have:$$ln y=10xlnleft(frac{14x}{14x+10}right)$$



And from here I get stuck trying to apply L'Hospital's Rule to find the limit.






calculus limits derivatives







share|cite|improve this question











$endgroup$




I am having a little trouble going about:
$$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x}$$



Using $ln$ properties we can bring down the $10x$ exponent and have:$$ln y=10xlnleft(frac{14x}{14x+10}right)$$



And from here I get stuck trying to apply L'Hospital's Rule to find the limit.






calculus limits derivatives




calculus limits derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 20:16









Bernard

119k740113




119k740113










asked Dec 4 '18 at 20:05









pijobordepijoborde

376




376












  • $begingroup$
    You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
    $endgroup$
    – gimusi
    Dec 4 '18 at 20:35


















  • $begingroup$
    You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
    $endgroup$
    – gimusi
    Dec 4 '18 at 20:35
















$begingroup$
You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
$endgroup$
– gimusi
Dec 4 '18 at 20:35




$begingroup$
You don't necessarly need l'Hopital here and I suggest to avoid that when we can proceed by standard limits.
$endgroup$
– gimusi
Dec 4 '18 at 20:35










5 Answers
5






active

oldest

votes


















4












$begingroup$

Hint :



$$10x ln left(frac{14x}{14x+10}right)= frac{ln left(frac{14x}{14x+10}right)}{frac{1}{10x}}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    HINT



    We can use that



    $$left(frac{14x}{14x+10}right)^{10x}=left(frac{14x+10-10}{14x+10}right)^{10x}=left[left(1-frac{10}{14x+10}right)^{frac{14x+10}{10}}right]^{frac{10cdot 10x}{14x+10}}$$



    and refer to standard limits.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      +1 by me since that was the original answer manipulating the definition of $e$.
      $endgroup$
      – Rebellos
      Dec 4 '18 at 21:42










    • $begingroup$
      @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
      $endgroup$
      – gimusi
      Dec 4 '18 at 21:45












    • $begingroup$
      I still don't get where e can come from here???
      $endgroup$
      – pijoborde
      Dec 4 '18 at 21:56










    • $begingroup$
      @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
      $endgroup$
      – gimusi
      Dec 4 '18 at 21:58












    • $begingroup$
      And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
      $endgroup$
      – pijoborde
      Dec 4 '18 at 22:02





















    3












    $begingroup$

    Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.



    $$ln y=10xlnleft(frac{14x}{14x+10}right)$$



    $$ln y=frac{lnleft(frac{14x}{14x+10}right)}{frac{1}{10x}}$$



    Now, you can continue.




    As an alternative approach (both faster and easier), you can use $lim_limits{n to infty}big(1+frac{x}{n}big)^n = e^x$. These expressions are easily manipulated to reach such a form.



    Notice the numerator is $10$ less than the denominator, so $frac{14x}{14x+14} = 1+frac{-10}{14x+10}$. Hence, you get



    $$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x} = lim_{xto infty} Biggl[left(1+frac{-10}{14x+10}right)^{14x+10}Biggl]^{frac{10x}{14x+10}}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) for the double hint
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:34










    • $begingroup$
      I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
      $endgroup$
      – pijoborde
      Dec 4 '18 at 20:45










    • $begingroup$
      @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:46










    • $begingroup$
      I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
      $endgroup$
      – KM101
      Dec 4 '18 at 20:47












    • $begingroup$
      Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
      $endgroup$
      – Rebellos
      Dec 4 '18 at 21:42





















    2












    $begingroup$

    Once you get to the logarithm, you need
    $$
    lim_{xtoinfty}frac{lndfrac{14x}{14x+10}}{dfrac{1}{10x}}
    $$
    which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $ln14+ln x-ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
    $$
    lim_{xtoinfty}frac{dfrac{1}{x}-dfrac{14}{14x+10}}{-dfrac{1}{10x^2}}=
    lim_{xtoinfty}-10x^2frac{14x+10-14x}{x(14x+10)}=lim_{xtoinfty}-frac{100x}{14x+10}
    $$
    When you have this limit, let me call it $l$, the one you started with is $e^l$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
      $endgroup$
      – pijoborde
      Dec 6 '18 at 7:01












    • $begingroup$
      @pijoborde $10cdot10=100$
      $endgroup$
      – egreg
      Dec 6 '18 at 9:01



















    1












    $begingroup$

    Tips:



    With equivalents, it's very fast:$$lnbiggl(frac{14x}{14x+10}biggr)=lnbiggl(1-frac{10}{14x+10}biggr)sim_infty -frac{10}{14x+10}sim_infty -frac{10}{14x}=-frac 57,$$
    so, as equivalence is compatible with multiplication/division,
    $$10xlnbiggl(frac{14x}{14x+10}biggr)sim_infty -frac{50x}{7x}to -frac{50}7.$$
    Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Equivalents are fast but also dangerous to handle for limits.
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:33












    • $begingroup$
      Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
      $endgroup$
      – Bernard
      Dec 4 '18 at 20:39








    • 1




      $begingroup$
      I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:44






    • 1




      $begingroup$
      I think you should always indicates that issue when you suggest to use equivalents to not expert users.
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:45






    • 2




      $begingroup$
      I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
      $endgroup$
      – gimusi
      Dec 4 '18 at 20:55













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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint :



    $$10x ln left(frac{14x}{14x+10}right)= frac{ln left(frac{14x}{14x+10}right)}{frac{1}{10x}}$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Hint :



      $$10x ln left(frac{14x}{14x+10}right)= frac{ln left(frac{14x}{14x+10}right)}{frac{1}{10x}}$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Hint :



        $$10x ln left(frac{14x}{14x+10}right)= frac{ln left(frac{14x}{14x+10}right)}{frac{1}{10x}}$$






        share|cite|improve this answer









        $endgroup$



        Hint :



        $$10x ln left(frac{14x}{14x+10}right)= frac{ln left(frac{14x}{14x+10}right)}{frac{1}{10x}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 20:07









        RebellosRebellos

        14.5k31246




        14.5k31246























            3












            $begingroup$

            HINT



            We can use that



            $$left(frac{14x}{14x+10}right)^{10x}=left(frac{14x+10-10}{14x+10}right)^{10x}=left[left(1-frac{10}{14x+10}right)^{frac{14x+10}{10}}right]^{frac{10cdot 10x}{14x+10}}$$



            and refer to standard limits.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 by me since that was the original answer manipulating the definition of $e$.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42










            • $begingroup$
              @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:45












            • $begingroup$
              I still don't get where e can come from here???
              $endgroup$
              – pijoborde
              Dec 4 '18 at 21:56










            • $begingroup$
              @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:58












            • $begingroup$
              And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
              $endgroup$
              – pijoborde
              Dec 4 '18 at 22:02


















            3












            $begingroup$

            HINT



            We can use that



            $$left(frac{14x}{14x+10}right)^{10x}=left(frac{14x+10-10}{14x+10}right)^{10x}=left[left(1-frac{10}{14x+10}right)^{frac{14x+10}{10}}right]^{frac{10cdot 10x}{14x+10}}$$



            and refer to standard limits.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 by me since that was the original answer manipulating the definition of $e$.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42










            • $begingroup$
              @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:45












            • $begingroup$
              I still don't get where e can come from here???
              $endgroup$
              – pijoborde
              Dec 4 '18 at 21:56










            • $begingroup$
              @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:58












            • $begingroup$
              And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
              $endgroup$
              – pijoborde
              Dec 4 '18 at 22:02
















            3












            3








            3





            $begingroup$

            HINT



            We can use that



            $$left(frac{14x}{14x+10}right)^{10x}=left(frac{14x+10-10}{14x+10}right)^{10x}=left[left(1-frac{10}{14x+10}right)^{frac{14x+10}{10}}right]^{frac{10cdot 10x}{14x+10}}$$



            and refer to standard limits.






            share|cite|improve this answer









            $endgroup$



            HINT



            We can use that



            $$left(frac{14x}{14x+10}right)^{10x}=left(frac{14x+10-10}{14x+10}right)^{10x}=left[left(1-frac{10}{14x+10}right)^{frac{14x+10}{10}}right]^{frac{10cdot 10x}{14x+10}}$$



            and refer to standard limits.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 20:06









            gimusigimusi

            92.9k94494




            92.9k94494












            • $begingroup$
              +1 by me since that was the original answer manipulating the definition of $e$.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42










            • $begingroup$
              @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:45












            • $begingroup$
              I still don't get where e can come from here???
              $endgroup$
              – pijoborde
              Dec 4 '18 at 21:56










            • $begingroup$
              @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:58












            • $begingroup$
              And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
              $endgroup$
              – pijoborde
              Dec 4 '18 at 22:02




















            • $begingroup$
              +1 by me since that was the original answer manipulating the definition of $e$.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42










            • $begingroup$
              @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:45












            • $begingroup$
              I still don't get where e can come from here???
              $endgroup$
              – pijoborde
              Dec 4 '18 at 21:56










            • $begingroup$
              @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
              $endgroup$
              – gimusi
              Dec 4 '18 at 21:58












            • $begingroup$
              And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
              $endgroup$
              – pijoborde
              Dec 4 '18 at 22:02


















            $begingroup$
            +1 by me since that was the original answer manipulating the definition of $e$.
            $endgroup$
            – Rebellos
            Dec 4 '18 at 21:42




            $begingroup$
            +1 by me since that was the original answer manipulating the definition of $e$.
            $endgroup$
            – Rebellos
            Dec 4 '18 at 21:42












            $begingroup$
            @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
            $endgroup$
            – gimusi
            Dec 4 '18 at 21:45






            $begingroup$
            @Rebellos Thanks also your hint is very effective since the asker was looking for a method to apply l'Hopital rule! :)
            $endgroup$
            – gimusi
            Dec 4 '18 at 21:45














            $begingroup$
            I still don't get where e can come from here???
            $endgroup$
            – pijoborde
            Dec 4 '18 at 21:56




            $begingroup$
            I still don't get where e can come from here???
            $endgroup$
            – pijoborde
            Dec 4 '18 at 21:56












            $begingroup$
            @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
            $endgroup$
            – gimusi
            Dec 4 '18 at 21:58






            $begingroup$
            @pijoborde Are you aware for as $nto infty$ $left(1+frac1nright)^n to e$ and $left(1+frac a nright)^n to e^a$?
            $endgroup$
            – gimusi
            Dec 4 '18 at 21:58














            $begingroup$
            And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
            $endgroup$
            – pijoborde
            Dec 4 '18 at 22:02






            $begingroup$
            And our a value is represented by 10 in this instance or? Making the answer e raised to 10? I mean, I feel like the answer is supposed to be e raised to some negative fraction.
            $endgroup$
            – pijoborde
            Dec 4 '18 at 22:02













            3












            $begingroup$

            Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.



            $$ln y=10xlnleft(frac{14x}{14x+10}right)$$



            $$ln y=frac{lnleft(frac{14x}{14x+10}right)}{frac{1}{10x}}$$



            Now, you can continue.




            As an alternative approach (both faster and easier), you can use $lim_limits{n to infty}big(1+frac{x}{n}big)^n = e^x$. These expressions are easily manipulated to reach such a form.



            Notice the numerator is $10$ less than the denominator, so $frac{14x}{14x+14} = 1+frac{-10}{14x+10}$. Hence, you get



            $$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x} = lim_{xto infty} Biggl[left(1+frac{-10}{14x+10}right)^{14x+10}Biggl]^{frac{10x}{14x+10}}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              (+1) for the double hint
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:34










            • $begingroup$
              I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
              $endgroup$
              – pijoborde
              Dec 4 '18 at 20:45










            • $begingroup$
              @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:46










            • $begingroup$
              I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
              $endgroup$
              – KM101
              Dec 4 '18 at 20:47












            • $begingroup$
              Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42


















            3












            $begingroup$

            Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.



            $$ln y=10xlnleft(frac{14x}{14x+10}right)$$



            $$ln y=frac{lnleft(frac{14x}{14x+10}right)}{frac{1}{10x}}$$



            Now, you can continue.




            As an alternative approach (both faster and easier), you can use $lim_limits{n to infty}big(1+frac{x}{n}big)^n = e^x$. These expressions are easily manipulated to reach such a form.



            Notice the numerator is $10$ less than the denominator, so $frac{14x}{14x+14} = 1+frac{-10}{14x+10}$. Hence, you get



            $$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x} = lim_{xto infty} Biggl[left(1+frac{-10}{14x+10}right)^{14x+10}Biggl]^{frac{10x}{14x+10}}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              (+1) for the double hint
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:34










            • $begingroup$
              I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
              $endgroup$
              – pijoborde
              Dec 4 '18 at 20:45










            • $begingroup$
              @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:46










            • $begingroup$
              I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
              $endgroup$
              – KM101
              Dec 4 '18 at 20:47












            • $begingroup$
              Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42
















            3












            3








            3





            $begingroup$

            Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.



            $$ln y=10xlnleft(frac{14x}{14x+10}right)$$



            $$ln y=frac{lnleft(frac{14x}{14x+10}right)}{frac{1}{10x}}$$



            Now, you can continue.




            As an alternative approach (both faster and easier), you can use $lim_limits{n to infty}big(1+frac{x}{n}big)^n = e^x$. These expressions are easily manipulated to reach such a form.



            Notice the numerator is $10$ less than the denominator, so $frac{14x}{14x+14} = 1+frac{-10}{14x+10}$. Hence, you get



            $$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x} = lim_{xto infty} Biggl[left(1+frac{-10}{14x+10}right)^{14x+10}Biggl]^{frac{10x}{14x+10}}$$






            share|cite|improve this answer











            $endgroup$



            Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.



            $$ln y=10xlnleft(frac{14x}{14x+10}right)$$



            $$ln y=frac{lnleft(frac{14x}{14x+10}right)}{frac{1}{10x}}$$



            Now, you can continue.




            As an alternative approach (both faster and easier), you can use $lim_limits{n to infty}big(1+frac{x}{n}big)^n = e^x$. These expressions are easily manipulated to reach such a form.



            Notice the numerator is $10$ less than the denominator, so $frac{14x}{14x+14} = 1+frac{-10}{14x+10}$. Hence, you get



            $$lim_{xto infty} left(frac{14x}{14x+10}right)^{10x} = lim_{xto infty} Biggl[left(1+frac{-10}{14x+10}right)^{14x+10}Biggl]^{frac{10x}{14x+10}}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 20:31

























            answered Dec 4 '18 at 20:26









            KM101KM101

            5,9131523




            5,9131523












            • $begingroup$
              (+1) for the double hint
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:34










            • $begingroup$
              I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
              $endgroup$
              – pijoborde
              Dec 4 '18 at 20:45










            • $begingroup$
              @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:46










            • $begingroup$
              I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
              $endgroup$
              – KM101
              Dec 4 '18 at 20:47












            • $begingroup$
              Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42




















            • $begingroup$
              (+1) for the double hint
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:34










            • $begingroup$
              I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
              $endgroup$
              – pijoborde
              Dec 4 '18 at 20:45










            • $begingroup$
              @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:46










            • $begingroup$
              I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
              $endgroup$
              – KM101
              Dec 4 '18 at 20:47












            • $begingroup$
              Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
              $endgroup$
              – Rebellos
              Dec 4 '18 at 21:42


















            $begingroup$
            (+1) for the double hint
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:34




            $begingroup$
            (+1) for the double hint
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:34












            $begingroup$
            I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
            $endgroup$
            – pijoborde
            Dec 4 '18 at 20:45




            $begingroup$
            I am supposed to get an answer involving e. So I need to focus on your alternative approach method?
            $endgroup$
            – pijoborde
            Dec 4 '18 at 20:45












            $begingroup$
            @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:46




            $begingroup$
            @pijoborde You can use both method. I prefer and suggest the second one which is much more instructive.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:46












            $begingroup$
            I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
            $endgroup$
            – KM101
            Dec 4 '18 at 20:47






            $begingroup$
            I don't know if you've worked on the limit provided for $e^x$. If you're allowed to use it, why not? For all limits like this, just look for a way to reach something in the form $$lim_{n to infty}bigg[bigg(1+frac{x}{n}bigg)^nbigg]^m$$ and evaluate the limit from there. It's very simple and has little room for error.
            $endgroup$
            – KM101
            Dec 4 '18 at 20:47














            $begingroup$
            Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
            $endgroup$
            – Rebellos
            Dec 4 '18 at 21:42






            $begingroup$
            Your answers are complete, but in my honest opinion, I don't find much sense in answering the same thing as others have first done and just elaborating it. The first part of your answer was answered by me and the second part by Gimusi.
            $endgroup$
            – Rebellos
            Dec 4 '18 at 21:42













            2












            $begingroup$

            Once you get to the logarithm, you need
            $$
            lim_{xtoinfty}frac{lndfrac{14x}{14x+10}}{dfrac{1}{10x}}
            $$
            which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $ln14+ln x-ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
            $$
            lim_{xtoinfty}frac{dfrac{1}{x}-dfrac{14}{14x+10}}{-dfrac{1}{10x^2}}=
            lim_{xtoinfty}-10x^2frac{14x+10-14x}{x(14x+10)}=lim_{xtoinfty}-frac{100x}{14x+10}
            $$
            When you have this limit, let me call it $l$, the one you started with is $e^l$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
              $endgroup$
              – pijoborde
              Dec 6 '18 at 7:01












            • $begingroup$
              @pijoborde $10cdot10=100$
              $endgroup$
              – egreg
              Dec 6 '18 at 9:01
















            2












            $begingroup$

            Once you get to the logarithm, you need
            $$
            lim_{xtoinfty}frac{lndfrac{14x}{14x+10}}{dfrac{1}{10x}}
            $$
            which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $ln14+ln x-ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
            $$
            lim_{xtoinfty}frac{dfrac{1}{x}-dfrac{14}{14x+10}}{-dfrac{1}{10x^2}}=
            lim_{xtoinfty}-10x^2frac{14x+10-14x}{x(14x+10)}=lim_{xtoinfty}-frac{100x}{14x+10}
            $$
            When you have this limit, let me call it $l$, the one you started with is $e^l$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
              $endgroup$
              – pijoborde
              Dec 6 '18 at 7:01












            • $begingroup$
              @pijoborde $10cdot10=100$
              $endgroup$
              – egreg
              Dec 6 '18 at 9:01














            2












            2








            2





            $begingroup$

            Once you get to the logarithm, you need
            $$
            lim_{xtoinfty}frac{lndfrac{14x}{14x+10}}{dfrac{1}{10x}}
            $$
            which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $ln14+ln x-ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
            $$
            lim_{xtoinfty}frac{dfrac{1}{x}-dfrac{14}{14x+10}}{-dfrac{1}{10x^2}}=
            lim_{xtoinfty}-10x^2frac{14x+10-14x}{x(14x+10)}=lim_{xtoinfty}-frac{100x}{14x+10}
            $$
            When you have this limit, let me call it $l$, the one you started with is $e^l$.






            share|cite|improve this answer









            $endgroup$



            Once you get to the logarithm, you need
            $$
            lim_{xtoinfty}frac{lndfrac{14x}{14x+10}}{dfrac{1}{10x}}
            $$
            which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $ln14+ln x-ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get
            $$
            lim_{xtoinfty}frac{dfrac{1}{x}-dfrac{14}{14x+10}}{-dfrac{1}{10x^2}}=
            lim_{xtoinfty}-10x^2frac{14x+10-14x}{x(14x+10)}=lim_{xtoinfty}-frac{100x}{14x+10}
            $$
            When you have this limit, let me call it $l$, the one you started with is $e^l$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 21:27









            egregegreg

            180k1485202




            180k1485202












            • $begingroup$
              What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
              $endgroup$
              – pijoborde
              Dec 6 '18 at 7:01












            • $begingroup$
              @pijoborde $10cdot10=100$
              $endgroup$
              – egreg
              Dec 6 '18 at 9:01


















            • $begingroup$
              What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
              $endgroup$
              – pijoborde
              Dec 6 '18 at 7:01












            • $begingroup$
              @pijoborde $10cdot10=100$
              $endgroup$
              – egreg
              Dec 6 '18 at 9:01
















            $begingroup$
            What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
            $endgroup$
            – pijoborde
            Dec 6 '18 at 7:01






            $begingroup$
            What happened exactly in the last step? I don't quite get where 100 came from? Unless you are reducing an x from both top and bottom to 1 and then distributing the -10x to the 10, since the positive and minus 14x will cancel out?
            $endgroup$
            – pijoborde
            Dec 6 '18 at 7:01














            $begingroup$
            @pijoborde $10cdot10=100$
            $endgroup$
            – egreg
            Dec 6 '18 at 9:01




            $begingroup$
            @pijoborde $10cdot10=100$
            $endgroup$
            – egreg
            Dec 6 '18 at 9:01











            1












            $begingroup$

            Tips:



            With equivalents, it's very fast:$$lnbiggl(frac{14x}{14x+10}biggr)=lnbiggl(1-frac{10}{14x+10}biggr)sim_infty -frac{10}{14x+10}sim_infty -frac{10}{14x}=-frac 57,$$
            so, as equivalence is compatible with multiplication/division,
            $$10xlnbiggl(frac{14x}{14x+10}biggr)sim_infty -frac{50x}{7x}to -frac{50}7.$$
            Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Equivalents are fast but also dangerous to handle for limits.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:33












            • $begingroup$
              Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
              $endgroup$
              – Bernard
              Dec 4 '18 at 20:39








            • 1




              $begingroup$
              I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:44






            • 1




              $begingroup$
              I think you should always indicates that issue when you suggest to use equivalents to not expert users.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:45






            • 2




              $begingroup$
              I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:55


















            1












            $begingroup$

            Tips:



            With equivalents, it's very fast:$$lnbiggl(frac{14x}{14x+10}biggr)=lnbiggl(1-frac{10}{14x+10}biggr)sim_infty -frac{10}{14x+10}sim_infty -frac{10}{14x}=-frac 57,$$
            so, as equivalence is compatible with multiplication/division,
            $$10xlnbiggl(frac{14x}{14x+10}biggr)sim_infty -frac{50x}{7x}to -frac{50}7.$$
            Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Equivalents are fast but also dangerous to handle for limits.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:33












            • $begingroup$
              Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
              $endgroup$
              – Bernard
              Dec 4 '18 at 20:39








            • 1




              $begingroup$
              I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:44






            • 1




              $begingroup$
              I think you should always indicates that issue when you suggest to use equivalents to not expert users.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:45






            • 2




              $begingroup$
              I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:55
















            1












            1








            1





            $begingroup$

            Tips:



            With equivalents, it's very fast:$$lnbiggl(frac{14x}{14x+10}biggr)=lnbiggl(1-frac{10}{14x+10}biggr)sim_infty -frac{10}{14x+10}sim_infty -frac{10}{14x}=-frac 57,$$
            so, as equivalence is compatible with multiplication/division,
            $$10xlnbiggl(frac{14x}{14x+10}biggr)sim_infty -frac{50x}{7x}to -frac{50}7.$$
            Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.






            share|cite|improve this answer











            $endgroup$



            Tips:



            With equivalents, it's very fast:$$lnbiggl(frac{14x}{14x+10}biggr)=lnbiggl(1-frac{10}{14x+10}biggr)sim_infty -frac{10}{14x+10}sim_infty -frac{10}{14x}=-frac 57,$$
            so, as equivalence is compatible with multiplication/division,
            $$10xlnbiggl(frac{14x}{14x+10}biggr)sim_infty -frac{50x}{7x}to -frac{50}7.$$
            Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 18:43

























            answered Dec 4 '18 at 20:25









            BernardBernard

            119k740113




            119k740113








            • 1




              $begingroup$
              Equivalents are fast but also dangerous to handle for limits.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:33












            • $begingroup$
              Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
              $endgroup$
              – Bernard
              Dec 4 '18 at 20:39








            • 1




              $begingroup$
              I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:44






            • 1




              $begingroup$
              I think you should always indicates that issue when you suggest to use equivalents to not expert users.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:45






            • 2




              $begingroup$
              I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:55
















            • 1




              $begingroup$
              Equivalents are fast but also dangerous to handle for limits.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:33












            • $begingroup$
              Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
              $endgroup$
              – Bernard
              Dec 4 '18 at 20:39








            • 1




              $begingroup$
              I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:44






            • 1




              $begingroup$
              I think you should always indicates that issue when you suggest to use equivalents to not expert users.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:45






            • 2




              $begingroup$
              I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
              $endgroup$
              – gimusi
              Dec 4 '18 at 20:55










            1




            1




            $begingroup$
            Equivalents are fast but also dangerous to handle for limits.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:33






            $begingroup$
            Equivalents are fast but also dangerous to handle for limits.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:33














            $begingroup$
            Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
            $endgroup$
            – Bernard
            Dec 4 '18 at 20:39






            $begingroup$
            Why? If one knows the rules… When I was a student, our professor used to say that L'Hospital's rule can be dangerous (there are some hypotheses no one ever thinks to check), and when it works, Taylor at order $1$ works as well.
            $endgroup$
            – Bernard
            Dec 4 '18 at 20:39






            1




            1




            $begingroup$
            I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:44




            $begingroup$
            I'm not referring of course to the use by an expert user like you but for not expert user it can be dangerous. Let consider for example the limit $frac{t-log(1+t)}{t^2}$ accorging to $log (1+t) sim t$ one could conclude that $$frac{t-log(1+t)}{t^2}sim frac{t-t}{t^2}=0$$ which is of course wrong.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:44




            1




            1




            $begingroup$
            I think you should always indicates that issue when you suggest to use equivalents to not expert users.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:45




            $begingroup$
            I think you should always indicates that issue when you suggest to use equivalents to not expert users.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:45




            2




            2




            $begingroup$
            I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:55






            $begingroup$
            I'm not claiming that the method is wrong, I'm claiming that we need to be aware about it to handle them and a not expert user can easily make a mistake. I would suggest for example use them in the form $log(1+t)=t+o(t)$ in order to take into account the rermainder term. But it is only a suggestion of course.
            $endgroup$
            – gimusi
            Dec 4 '18 at 20:55




















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