Limit expression using approximations












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$begingroup$


I have 2 questions that I'm unsure how to answer, they are both related to same question and function.



I am given a function:



$$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$



Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.



Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.



Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.



Thank you so much for your help in advance.






calculus limits approximation







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    0












    $begingroup$


    I have 2 questions that I'm unsure how to answer, they are both related to same question and function.



    I am given a function:



    $$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$



    Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.



    Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.



    Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.



    Thank you so much for your help in advance.






    calculus limits approximation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have 2 questions that I'm unsure how to answer, they are both related to same question and function.



      I am given a function:



      $$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$



      Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.



      Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.



      Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.



      Thank you so much for your help in advance.






      calculus limits approximation







      share|cite|improve this question











      $endgroup$




      I have 2 questions that I'm unsure how to answer, they are both related to same question and function.



      I am given a function:



      $$f(x)=ln(ln(1+sin^{2}x)+1)+cos (x) e^{sin x}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$



      Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.



      Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.



      Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.



      Thank you so much for your help in advance.






      calculus limits approximation




      calculus limits approximation






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 6:10









      Paramanand Singh

      51.3k558170




      51.3k558170










      asked Jan 5 at 23:38









      The StatisticianThe Statistician

      115112




      115112






















          1 Answer
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          $begingroup$

          We have that
          $$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
          and so $f(x)approx f(0)+xf'(0).$ Since
          $$f'(0) = frac{e-1}{e}$$
          and
          $$f(0) = 1$$
          we get that
          $$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$



          So for question $1$ if $g$ is an approximation of $f$ then it must be that
          $$f(x)approx f(0) + xf'(0)=1+x $$
          and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$



          For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
            $endgroup$
            – The Statistician
            Jan 6 at 0:58












          Your Answer





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          1 Answer
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          0












          $begingroup$

          We have that
          $$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
          and so $f(x)approx f(0)+xf'(0).$ Since
          $$f'(0) = frac{e-1}{e}$$
          and
          $$f(0) = 1$$
          we get that
          $$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$



          So for question $1$ if $g$ is an approximation of $f$ then it must be that
          $$f(x)approx f(0) + xf'(0)=1+x $$
          and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$



          For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
            $endgroup$
            – The Statistician
            Jan 6 at 0:58
















          0












          $begingroup$

          We have that
          $$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
          and so $f(x)approx f(0)+xf'(0).$ Since
          $$f'(0) = frac{e-1}{e}$$
          and
          $$f(0) = 1$$
          we get that
          $$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$



          So for question $1$ if $g$ is an approximation of $f$ then it must be that
          $$f(x)approx f(0) + xf'(0)=1+x $$
          and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$



          For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
            $endgroup$
            – The Statistician
            Jan 6 at 0:58














          0












          0








          0





          $begingroup$

          We have that
          $$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
          and so $f(x)approx f(0)+xf'(0).$ Since
          $$f'(0) = frac{e-1}{e}$$
          and
          $$f(0) = 1$$
          we get that
          $$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$



          So for question $1$ if $g$ is an approximation of $f$ then it must be that
          $$f(x)approx f(0) + xf'(0)=1+x $$
          and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$



          For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$






          share|cite|improve this answer









          $endgroup$



          We have that
          $$f(x)=ln(ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-frac{ln(x+1)}{e^{x^{2}+x+1}}$$
          and so $f(x)approx f(0)+xf'(0).$ Since
          $$f'(0) = frac{e-1}{e}$$
          and
          $$f(0) = 1$$
          we get that
          $$f(x)approx 1 + xleft(frac{e-1}{e}right) = 1+x-x/e.$$



          So for question $1$ if $g$ is an approximation of $f$ then it must be that
          $$f(x)approx f(0) + xf'(0)=1+x $$
          and so it must be the case that $$lim_{xto 0}frac{f(x)}{1+x}=1.$$



          For the second question, you can see that since $f'(0)neq 1$, $g(x)$ is not an approximation of $f$ near $0.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 0:05









          model_checkermodel_checker

          4,45521931




          4,45521931












          • $begingroup$
            Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
            $endgroup$
            – The Statistician
            Jan 6 at 0:58


















          • $begingroup$
            Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
            $endgroup$
            – The Statistician
            Jan 6 at 0:58
















          $begingroup$
          Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
          $endgroup$
          – The Statistician
          Jan 6 at 0:58




          $begingroup$
          Oh my gosh, thank you so much for this. This has helped me so much and it's so clear. Thank you.
          $endgroup$
          – The Statistician
          Jan 6 at 0:58


















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