Compute the limit $lim_{x to 0^+} (sin x)^{sin x}$ using L'Hopital's












0












$begingroup$


I needed some help with computing a limit of a function. I can compute it directly but I'm asked to use L'Hopital's and can't see where to start or how to use it for my function.



My limit expression is:



$$lim_{x to 0^+} (sin x)^{sin x}.$$



Any help would be appreciated. Thanks.






calculus limits







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$endgroup$








  • 3




    $begingroup$
    Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
    $endgroup$
    – T. Bongers
    Jan 5 at 0:40












  • $begingroup$
    As you now have it as a fraction, do i compute L'Hopitals on this?
    $endgroup$
    – The Statistician
    Jan 5 at 0:48






  • 1




    $begingroup$
    Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
    $endgroup$
    – Paramanand Singh
    Jan 5 at 12:55


















0












$begingroup$


I needed some help with computing a limit of a function. I can compute it directly but I'm asked to use L'Hopital's and can't see where to start or how to use it for my function.



My limit expression is:



$$lim_{x to 0^+} (sin x)^{sin x}.$$



Any help would be appreciated. Thanks.






calculus limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
    $endgroup$
    – T. Bongers
    Jan 5 at 0:40












  • $begingroup$
    As you now have it as a fraction, do i compute L'Hopitals on this?
    $endgroup$
    – The Statistician
    Jan 5 at 0:48






  • 1




    $begingroup$
    Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
    $endgroup$
    – Paramanand Singh
    Jan 5 at 12:55
















0












0








0





$begingroup$


I needed some help with computing a limit of a function. I can compute it directly but I'm asked to use L'Hopital's and can't see where to start or how to use it for my function.



My limit expression is:



$$lim_{x to 0^+} (sin x)^{sin x}.$$



Any help would be appreciated. Thanks.






calculus limits







share|cite|improve this question











$endgroup$




I needed some help with computing a limit of a function. I can compute it directly but I'm asked to use L'Hopital's and can't see where to start or how to use it for my function.



My limit expression is:



$$lim_{x to 0^+} (sin x)^{sin x}.$$



Any help would be appreciated. Thanks.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 0:40









T. Bongers

23.5k54762




23.5k54762










asked Jan 5 at 0:38









The StatisticianThe Statistician

115112




115112








  • 3




    $begingroup$
    Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
    $endgroup$
    – T. Bongers
    Jan 5 at 0:40












  • $begingroup$
    As you now have it as a fraction, do i compute L'Hopitals on this?
    $endgroup$
    – The Statistician
    Jan 5 at 0:48






  • 1




    $begingroup$
    Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
    $endgroup$
    – Paramanand Singh
    Jan 5 at 12:55
















  • 3




    $begingroup$
    Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
    $endgroup$
    – T. Bongers
    Jan 5 at 0:40












  • $begingroup$
    As you now have it as a fraction, do i compute L'Hopitals on this?
    $endgroup$
    – The Statistician
    Jan 5 at 0:48






  • 1




    $begingroup$
    Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
    $endgroup$
    – Paramanand Singh
    Jan 5 at 12:55










3




3




$begingroup$
Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
$endgroup$
– T. Bongers
Jan 5 at 0:40






$begingroup$
Take a logarithm first, and notice that $sin x ln sin x = frac{ln sin x}{1/sin x}$ becomes $$frac{cos x / sin x}{-cos x / sin^2 x}.$$
$endgroup$
– T. Bongers
Jan 5 at 0:40














$begingroup$
As you now have it as a fraction, do i compute L'Hopitals on this?
$endgroup$
– The Statistician
Jan 5 at 0:48




$begingroup$
As you now have it as a fraction, do i compute L'Hopitals on this?
$endgroup$
– The Statistician
Jan 5 at 0:48




1




1




$begingroup$
Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
$endgroup$
– Paramanand Singh
Jan 5 at 12:55






$begingroup$
Why not get rid of $sin x$ by using $t=sin x$ and then your limit is equal to $lim_{tto 0^{+}}t^t$.
$endgroup$
– Paramanand Singh
Jan 5 at 12:55












2 Answers
2






active

oldest

votes


















5












$begingroup$

Let's say the limit is $y$:



$$y=lim_{xto 0^+} (sin x)^{sin x}$$



Let's take the natural log of both sides in order to get rid of the exponentiation:



$$ln y=lim_{xto 0^+} (sin x)ln(sin x)$$



Now, this is a $0cdot -infty$ expression, which is indeterminate, but not a fraction where we can apply L'Hopital's rule. One really easy way to get this kind of expression into a fraction is by changing $sin x$ into $frac{1}{frac{1}{sin x}}$:



$$ln y=lim_{xto 0^+} frac{1}{frac{1}{sin x}}ln(sin x)=frac{ln(sin x)}{frac{1}{sin x}}$$



Now, we have a $frac{-infty}{infty}$ expression, so we can apply L'Hopital's. Take the derivative of both the numerator and denominator:



$$ln y=lim_{xto 0^+} frac{frac{cos x}{sin x}}{frac{-cos x}{sin^2 x}}$$



Simplify the fraction:



$$ln y=lim_{xto 0^+} -sin x=-sin 0=0$$



Thus, since $ln y=0$, we find that $y=1$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Using T.Bongers hint we are first of all left with the following




    $$log(L)=lim_{xto 0^+}sin(x)log(sin x)$$




    Approaching zero from the RHS we can conclude that the sine term tends to $0$ whereas the logarithmic term tends to negative infinity. Hence we are forced to establish the form $fracinftyinfty$ to use L'Hospitals rule we may rewrite the product as a fraction which fulfills the given condititons



    $$begin{align*}
    log(L)&=lim_{xto 0^+}sin(x)log(sin x)\
    &=lim_{xto 0^+}frac{log(sin x)}{1/sin(x)}\
    &=lim_{xto 0^+}frac{cot(x)}{-1cot(x)/sin(x)}\
    &=lim_{xto 0^+}-sin(x)\
    log(L)&=0
    end{align*}$$




    $$therefore~L=1$$







    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Let's say the limit is $y$:



      $$y=lim_{xto 0^+} (sin x)^{sin x}$$



      Let's take the natural log of both sides in order to get rid of the exponentiation:



      $$ln y=lim_{xto 0^+} (sin x)ln(sin x)$$



      Now, this is a $0cdot -infty$ expression, which is indeterminate, but not a fraction where we can apply L'Hopital's rule. One really easy way to get this kind of expression into a fraction is by changing $sin x$ into $frac{1}{frac{1}{sin x}}$:



      $$ln y=lim_{xto 0^+} frac{1}{frac{1}{sin x}}ln(sin x)=frac{ln(sin x)}{frac{1}{sin x}}$$



      Now, we have a $frac{-infty}{infty}$ expression, so we can apply L'Hopital's. Take the derivative of both the numerator and denominator:



      $$ln y=lim_{xto 0^+} frac{frac{cos x}{sin x}}{frac{-cos x}{sin^2 x}}$$



      Simplify the fraction:



      $$ln y=lim_{xto 0^+} -sin x=-sin 0=0$$



      Thus, since $ln y=0$, we find that $y=1$.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Let's say the limit is $y$:



        $$y=lim_{xto 0^+} (sin x)^{sin x}$$



        Let's take the natural log of both sides in order to get rid of the exponentiation:



        $$ln y=lim_{xto 0^+} (sin x)ln(sin x)$$



        Now, this is a $0cdot -infty$ expression, which is indeterminate, but not a fraction where we can apply L'Hopital's rule. One really easy way to get this kind of expression into a fraction is by changing $sin x$ into $frac{1}{frac{1}{sin x}}$:



        $$ln y=lim_{xto 0^+} frac{1}{frac{1}{sin x}}ln(sin x)=frac{ln(sin x)}{frac{1}{sin x}}$$



        Now, we have a $frac{-infty}{infty}$ expression, so we can apply L'Hopital's. Take the derivative of both the numerator and denominator:



        $$ln y=lim_{xto 0^+} frac{frac{cos x}{sin x}}{frac{-cos x}{sin^2 x}}$$



        Simplify the fraction:



        $$ln y=lim_{xto 0^+} -sin x=-sin 0=0$$



        Thus, since $ln y=0$, we find that $y=1$.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Let's say the limit is $y$:



          $$y=lim_{xto 0^+} (sin x)^{sin x}$$



          Let's take the natural log of both sides in order to get rid of the exponentiation:



          $$ln y=lim_{xto 0^+} (sin x)ln(sin x)$$



          Now, this is a $0cdot -infty$ expression, which is indeterminate, but not a fraction where we can apply L'Hopital's rule. One really easy way to get this kind of expression into a fraction is by changing $sin x$ into $frac{1}{frac{1}{sin x}}$:



          $$ln y=lim_{xto 0^+} frac{1}{frac{1}{sin x}}ln(sin x)=frac{ln(sin x)}{frac{1}{sin x}}$$



          Now, we have a $frac{-infty}{infty}$ expression, so we can apply L'Hopital's. Take the derivative of both the numerator and denominator:



          $$ln y=lim_{xto 0^+} frac{frac{cos x}{sin x}}{frac{-cos x}{sin^2 x}}$$



          Simplify the fraction:



          $$ln y=lim_{xto 0^+} -sin x=-sin 0=0$$



          Thus, since $ln y=0$, we find that $y=1$.






          share|cite|improve this answer









          $endgroup$



          Let's say the limit is $y$:



          $$y=lim_{xto 0^+} (sin x)^{sin x}$$



          Let's take the natural log of both sides in order to get rid of the exponentiation:



          $$ln y=lim_{xto 0^+} (sin x)ln(sin x)$$



          Now, this is a $0cdot -infty$ expression, which is indeterminate, but not a fraction where we can apply L'Hopital's rule. One really easy way to get this kind of expression into a fraction is by changing $sin x$ into $frac{1}{frac{1}{sin x}}$:



          $$ln y=lim_{xto 0^+} frac{1}{frac{1}{sin x}}ln(sin x)=frac{ln(sin x)}{frac{1}{sin x}}$$



          Now, we have a $frac{-infty}{infty}$ expression, so we can apply L'Hopital's. Take the derivative of both the numerator and denominator:



          $$ln y=lim_{xto 0^+} frac{frac{cos x}{sin x}}{frac{-cos x}{sin^2 x}}$$



          Simplify the fraction:



          $$ln y=lim_{xto 0^+} -sin x=-sin 0=0$$



          Thus, since $ln y=0$, we find that $y=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 0:54









          Noble MushtakNoble Mushtak

          15.3k1835




          15.3k1835























              1












              $begingroup$

              Using T.Bongers hint we are first of all left with the following




              $$log(L)=lim_{xto 0^+}sin(x)log(sin x)$$




              Approaching zero from the RHS we can conclude that the sine term tends to $0$ whereas the logarithmic term tends to negative infinity. Hence we are forced to establish the form $fracinftyinfty$ to use L'Hospitals rule we may rewrite the product as a fraction which fulfills the given condititons



              $$begin{align*}
              log(L)&=lim_{xto 0^+}sin(x)log(sin x)\
              &=lim_{xto 0^+}frac{log(sin x)}{1/sin(x)}\
              &=lim_{xto 0^+}frac{cot(x)}{-1cot(x)/sin(x)}\
              &=lim_{xto 0^+}-sin(x)\
              log(L)&=0
              end{align*}$$




              $$therefore~L=1$$







              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Using T.Bongers hint we are first of all left with the following




                $$log(L)=lim_{xto 0^+}sin(x)log(sin x)$$




                Approaching zero from the RHS we can conclude that the sine term tends to $0$ whereas the logarithmic term tends to negative infinity. Hence we are forced to establish the form $fracinftyinfty$ to use L'Hospitals rule we may rewrite the product as a fraction which fulfills the given condititons



                $$begin{align*}
                log(L)&=lim_{xto 0^+}sin(x)log(sin x)\
                &=lim_{xto 0^+}frac{log(sin x)}{1/sin(x)}\
                &=lim_{xto 0^+}frac{cot(x)}{-1cot(x)/sin(x)}\
                &=lim_{xto 0^+}-sin(x)\
                log(L)&=0
                end{align*}$$




                $$therefore~L=1$$







                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Using T.Bongers hint we are first of all left with the following




                  $$log(L)=lim_{xto 0^+}sin(x)log(sin x)$$




                  Approaching zero from the RHS we can conclude that the sine term tends to $0$ whereas the logarithmic term tends to negative infinity. Hence we are forced to establish the form $fracinftyinfty$ to use L'Hospitals rule we may rewrite the product as a fraction which fulfills the given condititons



                  $$begin{align*}
                  log(L)&=lim_{xto 0^+}sin(x)log(sin x)\
                  &=lim_{xto 0^+}frac{log(sin x)}{1/sin(x)}\
                  &=lim_{xto 0^+}frac{cot(x)}{-1cot(x)/sin(x)}\
                  &=lim_{xto 0^+}-sin(x)\
                  log(L)&=0
                  end{align*}$$




                  $$therefore~L=1$$







                  share|cite|improve this answer









                  $endgroup$



                  Using T.Bongers hint we are first of all left with the following




                  $$log(L)=lim_{xto 0^+}sin(x)log(sin x)$$




                  Approaching zero from the RHS we can conclude that the sine term tends to $0$ whereas the logarithmic term tends to negative infinity. Hence we are forced to establish the form $fracinftyinfty$ to use L'Hospitals rule we may rewrite the product as a fraction which fulfills the given condititons



                  $$begin{align*}
                  log(L)&=lim_{xto 0^+}sin(x)log(sin x)\
                  &=lim_{xto 0^+}frac{log(sin x)}{1/sin(x)}\
                  &=lim_{xto 0^+}frac{cot(x)}{-1cot(x)/sin(x)}\
                  &=lim_{xto 0^+}-sin(x)\
                  log(L)&=0
                  end{align*}$$




                  $$therefore~L=1$$








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 0:55









                  mrtaurhomrtaurho

                  6,11271641




                  6,11271641






























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