Limit using the definition of a convergent sequence












-1












$begingroup$


I am trying to show that $lim_{nto+infty}frac{3n-5}{2n-1} = frac{3}{2}$ by using the definition of a convergent sequence. so far I have,
$lvert a_n - arvert < epsilon$, for any $epsilon > 0 $



$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $



$lvert frac{-7}{4n-2}rvert < epsilon$



$ frac{7}{4n-2} < epsilon$



But I am struggling to find a way to compare it to a sequence without the -2 in the denominator or how to manipulate it. Any help would be great, Thank you.






sequences-and-series analysis







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    I am trying to show that $lim_{nto+infty}frac{3n-5}{2n-1} = frac{3}{2}$ by using the definition of a convergent sequence. so far I have,
    $lvert a_n - arvert < epsilon$, for any $epsilon > 0 $



    $lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $



    $lvert frac{-7}{4n-2}rvert < epsilon$



    $ frac{7}{4n-2} < epsilon$



    But I am struggling to find a way to compare it to a sequence without the -2 in the denominator or how to manipulate it. Any help would be great, Thank you.






    sequences-and-series analysis







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I am trying to show that $lim_{nto+infty}frac{3n-5}{2n-1} = frac{3}{2}$ by using the definition of a convergent sequence. so far I have,
      $lvert a_n - arvert < epsilon$, for any $epsilon > 0 $



      $lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $



      $lvert frac{-7}{4n-2}rvert < epsilon$



      $ frac{7}{4n-2} < epsilon$



      But I am struggling to find a way to compare it to a sequence without the -2 in the denominator or how to manipulate it. Any help would be great, Thank you.






      sequences-and-series analysis







      share|cite|improve this question











      $endgroup$




      I am trying to show that $lim_{nto+infty}frac{3n-5}{2n-1} = frac{3}{2}$ by using the definition of a convergent sequence. so far I have,
      $lvert a_n - arvert < epsilon$, for any $epsilon > 0 $



      $lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $



      $lvert frac{-7}{4n-2}rvert < epsilon$



      $ frac{7}{4n-2} < epsilon$



      But I am struggling to find a way to compare it to a sequence without the -2 in the denominator or how to manipulate it. Any help would be great, Thank you.






      sequences-and-series analysis




      sequences-and-series analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 16:30







      adsy9

















      asked Dec 3 '18 at 16:18









      adsy9adsy9

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          3 Answers
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          $begingroup$

          $$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $$



          $$lvert frac{-7}{4n-2}rvert color{red}< epsilon$$



          $$ frac{7}{4n-2} < epsilon$$



          $$4n-2 > frac7epsilon$$



          $$n> frac14left( frac7epsilon + 2 right)$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You have made a mistake there



            $$left| frac{3n-5}{2n-1} - frac{3}{2} right| < epsilon implies left| frac{-7}{4n-2} right| < epsilon$$



            and now it is easy conclude form here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oh yea, sorry, I put the wrong inequality.
              $endgroup$
              – adsy9
              Dec 3 '18 at 16:31










            • $begingroup$
              @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
              $endgroup$
              – gimusi
              Dec 3 '18 at 16:39



















            -1












            $begingroup$

            Something of importance to note here is that when you are working on finding the $epsilon$'s and $N$'s you want to use in the proof you can begin by assuming that $|a_n-a|<epsilon$, but when you go to write the proof you must not start that way.



            So, you want to try and find an $n$ so that $frac{7}{4n-2}<epsilon$. Well, I don't like that fraction. It confuses me. So let's replace it with another fraction that is nicer! We know that $frac{7}{4n-2}<frac{7}{n}$ for all $n$ (though we don't technically need for that to be the case). Then, so long as $frac{7}{n}<epsilon$ we will be just as good as before.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
              $endgroup$
              – ImNotTheSaxMan
              Dec 3 '18 at 16:26











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            3 Answers
            3






            active

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            3 Answers
            3






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            active

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            1












            $begingroup$

            $$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $$



            $$lvert frac{-7}{4n-2}rvert color{red}< epsilon$$



            $$ frac{7}{4n-2} < epsilon$$



            $$4n-2 > frac7epsilon$$



            $$n> frac14left( frac7epsilon + 2 right)$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $$



              $$lvert frac{-7}{4n-2}rvert color{red}< epsilon$$



              $$ frac{7}{4n-2} < epsilon$$



              $$4n-2 > frac7epsilon$$



              $$n> frac14left( frac7epsilon + 2 right)$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $$



                $$lvert frac{-7}{4n-2}rvert color{red}< epsilon$$



                $$ frac{7}{4n-2} < epsilon$$



                $$4n-2 > frac7epsilon$$



                $$n> frac14left( frac7epsilon + 2 right)$$






                share|cite|improve this answer









                $endgroup$



                $$lvert frac{3n-5}{2n-1} - frac{3}{2} rvert < epsilon $$



                $$lvert frac{-7}{4n-2}rvert color{red}< epsilon$$



                $$ frac{7}{4n-2} < epsilon$$



                $$4n-2 > frac7epsilon$$



                $$n> frac14left( frac7epsilon + 2 right)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 16:22









                Siong Thye GohSiong Thye Goh

                100k1465117




                100k1465117























                    1












                    $begingroup$

                    You have made a mistake there



                    $$left| frac{3n-5}{2n-1} - frac{3}{2} right| < epsilon implies left| frac{-7}{4n-2} right| < epsilon$$



                    and now it is easy conclude form here.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Oh yea, sorry, I put the wrong inequality.
                      $endgroup$
                      – adsy9
                      Dec 3 '18 at 16:31










                    • $begingroup$
                      @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                      $endgroup$
                      – gimusi
                      Dec 3 '18 at 16:39
















                    1












                    $begingroup$

                    You have made a mistake there



                    $$left| frac{3n-5}{2n-1} - frac{3}{2} right| < epsilon implies left| frac{-7}{4n-2} right| < epsilon$$



                    and now it is easy conclude form here.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Oh yea, sorry, I put the wrong inequality.
                      $endgroup$
                      – adsy9
                      Dec 3 '18 at 16:31










                    • $begingroup$
                      @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                      $endgroup$
                      – gimusi
                      Dec 3 '18 at 16:39














                    1












                    1








                    1





                    $begingroup$

                    You have made a mistake there



                    $$left| frac{3n-5}{2n-1} - frac{3}{2} right| < epsilon implies left| frac{-7}{4n-2} right| < epsilon$$



                    and now it is easy conclude form here.






                    share|cite|improve this answer









                    $endgroup$



                    You have made a mistake there



                    $$left| frac{3n-5}{2n-1} - frac{3}{2} right| < epsilon implies left| frac{-7}{4n-2} right| < epsilon$$



                    and now it is easy conclude form here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 16:25









                    gimusigimusi

                    1




                    1












                    • $begingroup$
                      Oh yea, sorry, I put the wrong inequality.
                      $endgroup$
                      – adsy9
                      Dec 3 '18 at 16:31










                    • $begingroup$
                      @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                      $endgroup$
                      – gimusi
                      Dec 3 '18 at 16:39


















                    • $begingroup$
                      Oh yea, sorry, I put the wrong inequality.
                      $endgroup$
                      – adsy9
                      Dec 3 '18 at 16:31










                    • $begingroup$
                      @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                      $endgroup$
                      – gimusi
                      Dec 3 '18 at 16:39
















                    $begingroup$
                    Oh yea, sorry, I put the wrong inequality.
                    $endgroup$
                    – adsy9
                    Dec 3 '18 at 16:31




                    $begingroup$
                    Oh yea, sorry, I put the wrong inequality.
                    $endgroup$
                    – adsy9
                    Dec 3 '18 at 16:31












                    $begingroup$
                    @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                    $endgroup$
                    – gimusi
                    Dec 3 '18 at 16:39




                    $begingroup$
                    @adsy9 Then you are almost done form $frac{7}{4n-2} < epsilon$ assuming $4n-2>0$ we obtain $epsilon(4n-2)>7$ and now it suffices to isolate $n$.
                    $endgroup$
                    – gimusi
                    Dec 3 '18 at 16:39











                    -1












                    $begingroup$

                    Something of importance to note here is that when you are working on finding the $epsilon$'s and $N$'s you want to use in the proof you can begin by assuming that $|a_n-a|<epsilon$, but when you go to write the proof you must not start that way.



                    So, you want to try and find an $n$ so that $frac{7}{4n-2}<epsilon$. Well, I don't like that fraction. It confuses me. So let's replace it with another fraction that is nicer! We know that $frac{7}{4n-2}<frac{7}{n}$ for all $n$ (though we don't technically need for that to be the case). Then, so long as $frac{7}{n}<epsilon$ we will be just as good as before.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                      $endgroup$
                      – ImNotTheSaxMan
                      Dec 3 '18 at 16:26
















                    -1












                    $begingroup$

                    Something of importance to note here is that when you are working on finding the $epsilon$'s and $N$'s you want to use in the proof you can begin by assuming that $|a_n-a|<epsilon$, but when you go to write the proof you must not start that way.



                    So, you want to try and find an $n$ so that $frac{7}{4n-2}<epsilon$. Well, I don't like that fraction. It confuses me. So let's replace it with another fraction that is nicer! We know that $frac{7}{4n-2}<frac{7}{n}$ for all $n$ (though we don't technically need for that to be the case). Then, so long as $frac{7}{n}<epsilon$ we will be just as good as before.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                      $endgroup$
                      – ImNotTheSaxMan
                      Dec 3 '18 at 16:26














                    -1












                    -1








                    -1





                    $begingroup$

                    Something of importance to note here is that when you are working on finding the $epsilon$'s and $N$'s you want to use in the proof you can begin by assuming that $|a_n-a|<epsilon$, but when you go to write the proof you must not start that way.



                    So, you want to try and find an $n$ so that $frac{7}{4n-2}<epsilon$. Well, I don't like that fraction. It confuses me. So let's replace it with another fraction that is nicer! We know that $frac{7}{4n-2}<frac{7}{n}$ for all $n$ (though we don't technically need for that to be the case). Then, so long as $frac{7}{n}<epsilon$ we will be just as good as before.






                    share|cite|improve this answer









                    $endgroup$



                    Something of importance to note here is that when you are working on finding the $epsilon$'s and $N$'s you want to use in the proof you can begin by assuming that $|a_n-a|<epsilon$, but when you go to write the proof you must not start that way.



                    So, you want to try and find an $n$ so that $frac{7}{4n-2}<epsilon$. Well, I don't like that fraction. It confuses me. So let's replace it with another fraction that is nicer! We know that $frac{7}{4n-2}<frac{7}{n}$ for all $n$ (though we don't technically need for that to be the case). Then, so long as $frac{7}{n}<epsilon$ we will be just as good as before.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 16:24









                    ImNotTheSaxManImNotTheSaxMan

                    1172




                    1172












                    • $begingroup$
                      Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                      $endgroup$
                      – ImNotTheSaxMan
                      Dec 3 '18 at 16:26


















                    • $begingroup$
                      Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                      $endgroup$
                      – ImNotTheSaxMan
                      Dec 3 '18 at 16:26
















                    $begingroup$
                    Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                    $endgroup$
                    – ImNotTheSaxMan
                    Dec 3 '18 at 16:26




                    $begingroup$
                    Is there a particular reason this received a downvote? It addresses the problem in the OP in a way that generalizes to other problems.
                    $endgroup$
                    – ImNotTheSaxMan
                    Dec 3 '18 at 16:26


















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