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












-1














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 '18 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 '18 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 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 '18 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 at 11:49
















-1














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 '18 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 '18 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 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 '18 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 at 11:49














-1












-1








-1







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 '18 at 11:48









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Nov 22 '18 at 11:40









Rumman Bin Rayhan RijviRumman Bin Rayhan Rijvi

71




71




closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 '18 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 '18 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 '18 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 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 '18 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 at 11:49


















  • Does the limit exist?
    – Akash Roy
    Nov 22 '18 at 11:42










  • H Hospitals law limits need to solve this math
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 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 '18 at 11:47










  • L. hospitals law
    – Rumman Bin Rayhan Rijvi
    Nov 22 '18 at 11:49
















Does the limit exist?
– Akash Roy
Nov 22 '18 at 11:42




Does the limit exist?
– Akash Roy
Nov 22 '18 at 11:42












H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 '18 at 11:43




H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 '18 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 '18 at 11:47




No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 '18 at 11:47












L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 '18 at 11:49




L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 '18 at 11:49










2 Answers
2






active

oldest

votes


















2














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





























    1














    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









      2














      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


























        2














        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
























          2












          2








          2






          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 '18 at 11:47









          gimusigimusi

          1




          1























              1














              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


























                1














                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
























                  1












                  1








                  1






                  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 '18 at 12:22









                  YiFanYiFan

                  2,6941422




                  2,6941422















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