Compute $lim_{xto0} (sin x)^x$ using L'Hospital's rule [closed]











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Help me to compute this limit with L'Hospital's rule.



$$lim_{xto 0} (sin x)^x$$



Thanks in advance.






limits







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closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Does the limit exist?
    – Akash Roy
    Nov 22 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:43






  • 4




    No need to comment on someone's grammar mistakes, especially when English are not they're first language.
    – Rebellos
    Nov 22 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:49















up vote
-1
down vote

favorite












Help me to compute this limit with L'Hospital's rule.



$$lim_{xto 0} (sin x)^x$$



Thanks in advance.






limits







share|cite|improve this question















closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Does the limit exist?
    – Akash Roy
    Nov 22 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:43






  • 4




    No need to comment on someone's grammar mistakes, especially when English are not they're first language.
    – Rebellos
    Nov 22 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:49













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Help me to compute this limit with L'Hospital's rule.



$$lim_{xto 0} (sin x)^x$$



Thanks in advance.






limits







share|cite|improve this question















Help me to compute this limit with L'Hospital's rule.



$$lim_{xto 0} (sin x)^x$$



Thanks in advance.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 11:48









Jean-Claude Arbaut

14.7k63363




14.7k63363










asked Nov 22 at 11:40









Rumman Bin Rayhan Rijvi

71




71




closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Does the limit exist?
    – Akash Roy
    Nov 22 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:43






  • 4




    No need to comment on someone's grammar mistakes, especially when English are not they're first language.
    – Rebellos
    Nov 22 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:49


















  • Does the limit exist?
    – Akash Roy
    Nov 22 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:43






  • 4




    No need to comment on someone's grammar mistakes, especially when English are not they're first language.
    – Rebellos
    Nov 22 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 at 11:49
















Does the limit exist?
– Akash Roy
Nov 22 at 11:42




Does the limit exist?
– Akash Roy
Nov 22 at 11:42












H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43




H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43




4




4




No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47




No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47












L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49




L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49










2 Answers
2






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up vote
2
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HINT



We need $x>0$, then use



$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$






share|cite|improve this answer




























    up vote
    0
    down vote













    Use L'Hospital's rule to complete the standard exercise of finding that
    $$ lim_{xto0}frac{sin x}{x}=1.$$
    So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
    $$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
    Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      HINT



      We need $x>0$, then use



      $$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$






      share|cite|improve this answer

























        up vote
        2
        down vote













        HINT



        We need $x>0$, then use



        $$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          HINT



          We need $x>0$, then use



          $$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$






          share|cite|improve this answer












          HINT



          We need $x>0$, then use



          $$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 11:47









          gimusi

          92.5k94495




          92.5k94495






















              up vote
              0
              down vote













              Use L'Hospital's rule to complete the standard exercise of finding that
              $$ lim_{xto0}frac{sin x}{x}=1.$$
              So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
              $$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
              Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Use L'Hospital's rule to complete the standard exercise of finding that
                $$ lim_{xto0}frac{sin x}{x}=1.$$
                So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
                $$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
                Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Use L'Hospital's rule to complete the standard exercise of finding that
                  $$ lim_{xto0}frac{sin x}{x}=1.$$
                  So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
                  $$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
                  Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.






                  share|cite|improve this answer












                  Use L'Hospital's rule to complete the standard exercise of finding that
                  $$ lim_{xto0}frac{sin x}{x}=1.$$
                  So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
                  $$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
                  Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 12:22









                  YiFan

                  2,1761421




                  2,1761421















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