Limit without L'Hospital rule












0















The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$




Try: Put $displaystyle y=frac{1}{z}$



So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$



$$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$



Using L'Hospital rule



$$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$



Could some help me? How can I solve without L'Hospital rule? Thanks






limits limits-without-lhopital







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    0















    The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$




    Try: Put $displaystyle y=frac{1}{z}$



    So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$



    $$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$



    Using L'Hospital rule



    $$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$



    Could some help me? How can I solve without L'Hospital rule? Thanks






    limits limits-without-lhopital







    share|cite|improve this question



























      0












      0








      0








      The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$




      Try: Put $displaystyle y=frac{1}{z}$



      So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$



      $$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$



      Using L'Hospital rule



      $$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$



      Could some help me? How can I solve without L'Hospital rule? Thanks






      limits limits-without-lhopital







      share|cite|improve this question
















      The value of $$lim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg), xin bigg(0,frac{pi}{2}bigg)$$




      Try: Put $displaystyle y=frac{1}{z}$



      So $$lim_{zrightarrow 0}lnbigg(frac{sin(x+z)}{sin x}bigg)cdot frac{1}{z}$$



      $$ = lim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$$



      Using L'Hospital rule



      $$lim_{zrightarrow 0}frac{cos(x+z)}{sin (x+z)}=cot x$$



      Could some help me? How can I solve without L'Hospital rule? Thanks






      limits limits-without-lhopital




      limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited yesterday









      MJD

      46.9k28209392




      46.9k28209392










      asked 2 days ago









      D Tiwari

      5,3782630




      5,3782630






















          4 Answers
          4






          active

          oldest

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          3














          $displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$



          Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get



          $=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$






          share|cite|improve this answer























          • This is how I would have answered but I guess I was late. +1
            – Paramanand Singh
            yesterday



















          2














          $displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.






          share|cite|improve this answer

















          • 1




            Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
            – mathcounterexamples.net
            2 days ago












          • @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
            – Yadati Kiran
            2 days ago










          • @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
            – Paramanand Singh
            yesterday












          • You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
            – Paramanand Singh
            yesterday










          • @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
            – Yadati Kiran
            yesterday





















          2














          Beside the simple solution given by Yadati Kiran, you could use Taylor series
          $$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
          $$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
          $$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
          $$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.






          share|cite|improve this answer





























            1














            This is not a solution, more of a question.



            I get to



            1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$



            $y rightarrow infty : $



            Known:



            2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.



            (Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)



            Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?



            Could not come up with a solution.



            Thanks






            share|cite|improve this answer























            • The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
              – Paramanand Singh
              yesterday












            • Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
              – Peter Szilas
              yesterday











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            4 Answers
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            active

            oldest

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            4 Answers
            4






            active

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            active

            oldest

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            active

            oldest

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            3














            $displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$



            Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get



            $=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$






            share|cite|improve this answer























            • This is how I would have answered but I guess I was late. +1
              – Paramanand Singh
              yesterday
















            3














            $displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$



            Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get



            $=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$






            share|cite|improve this answer























            • This is how I would have answered but I guess I was late. +1
              – Paramanand Singh
              yesterday














            3












            3








            3






            $displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$



            Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get



            $=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$






            share|cite|improve this answer














            $displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x+1/y)}{sin x}bigg)=displaystylelim_{yrightarrow infty}y lnbigg(frac{sin(x)cos(1/y)+cos(x)sin(1/y)}{sin x}bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(cos(1/y)+cot(x)sin(1/y)bigg)\=displaystylelim_{yrightarrow infty}y lnbigg(1+big[cos(1/y)+cot(x)sin(1/y)-1big]bigg)$



            Since $cos(1/y)+cot(x)sin(1/y)-1to0$ as $ytoinfty$, using the standard limit $displaystylelim_{mto0}frac{ln(1+m)}{m}=1$, we get



            $=displaystylelim_{yrightarrow infty}ybig(cos(1/y)+cot(x)sin(1/y)-1big)\=displaystylelim_{zto0^+}frac{cos z-1}z+cot(x)cdotlim_{zto0^+}frac{sin z}z$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            Shubham Johri

            4,264717




            4,264717












            • This is how I would have answered but I guess I was late. +1
              – Paramanand Singh
              yesterday


















            • This is how I would have answered but I guess I was late. +1
              – Paramanand Singh
              yesterday
















            This is how I would have answered but I guess I was late. +1
            – Paramanand Singh
            yesterday




            This is how I would have answered but I guess I was late. +1
            – Paramanand Singh
            yesterday











            2














            $displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.






            share|cite|improve this answer

















            • 1




              Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
              – mathcounterexamples.net
              2 days ago












            • @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
              – Yadati Kiran
              2 days ago










            • @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
              – Paramanand Singh
              yesterday












            • You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
              – Paramanand Singh
              yesterday










            • @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
              – Yadati Kiran
              yesterday


















            2














            $displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.






            share|cite|improve this answer

















            • 1




              Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
              – mathcounterexamples.net
              2 days ago












            • @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
              – Yadati Kiran
              2 days ago










            • @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
              – Paramanand Singh
              yesterday












            • You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
              – Paramanand Singh
              yesterday










            • @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
              – Yadati Kiran
              yesterday
















            2












            2








            2






            $displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.






            share|cite|improve this answer












            $displaystylelim_{zrightarrow 0}frac{ln(sin (x+z))-ln(sin x)}{z}$ is precisely the derivative of $ln(sin x)$ for $xinleft(0,dfrac{pi}{2}right)$using limits.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Yadati Kiran

            1,694619




            1,694619








            • 1




              Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
              – mathcounterexamples.net
              2 days ago












            • @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
              – Yadati Kiran
              2 days ago










            • @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
              – Paramanand Singh
              yesterday












            • You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
              – Paramanand Singh
              yesterday










            • @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
              – Yadati Kiran
              yesterday
















            • 1




              Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
              – mathcounterexamples.net
              2 days ago












            • @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
              – Yadati Kiran
              2 days ago










            • @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
              – Paramanand Singh
              yesterday












            • You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
              – Paramanand Singh
              yesterday










            • @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
              – Yadati Kiran
              yesterday










            1




            1




            Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
            – mathcounterexamples.net
            2 days ago






            Using the derivative is precisely using L'Hôpital's rule. Which the OP is requesting not to do.
            – mathcounterexamples.net
            2 days ago














            @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
            – Yadati Kiran
            2 days ago




            @mathcounterexamples.net: You've got to use L'Hopital's rule at some stage.
            – Yadati Kiran
            2 days ago












            @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
            – Paramanand Singh
            yesterday






            @mathcounterexamples.net: Not exactly (L'Hospital's Rule involves differentiation followed by a limit operation) but if one forbids L'Hospital's Rule then the implicit intention is to forbid derivatives as well.
            – Paramanand Singh
            yesterday














            You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
            – Paramanand Singh
            yesterday




            You don't necessarily need to use L'Hospital's Rule at some stage. Just use basic limit laws and you get the desired answer.
            – Paramanand Singh
            yesterday












            @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
            – Yadati Kiran
            yesterday






            @ParamanandSingh: Differentiation with limits uses L'Hopital rule implicitly in certain cases. How will one evaluate using just basic limit laws ? Can you elaborate.
            – Yadati Kiran
            yesterday













            2














            Beside the simple solution given by Yadati Kiran, you could use Taylor series
            $$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
            $$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
            $$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
            $$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.






            share|cite|improve this answer


























              2














              Beside the simple solution given by Yadati Kiran, you could use Taylor series
              $$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
              $$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
              $$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
              $$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.






              share|cite|improve this answer
























                2












                2








                2






                Beside the simple solution given by Yadati Kiran, you could use Taylor series
                $$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
                $$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
                $$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
                $$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.






                share|cite|improve this answer












                Beside the simple solution given by Yadati Kiran, you could use Taylor series
                $$frac{sin(x+z)}{sin (x)}=cot (x) sin (z)+cos (z)$$ Now, using the classical expansion of $sin(z)$ and $cos(z)$ you then have
                $$cot (x) sin (z)+cos (z)=1+z cot (x)-frac{z^2}{2}+Oleft(z^3right)$$ Continue with the logarithm to get
                $$log (cot (x) sin (z)+cos (z))=z cot (x)-frac{1}{2} z^2 left(cot ^2(x)+1right)+Oleft(z^3right)$$
                $$frac 1 z log (cot (x) sin (z)+cos (z))= cot (x)-frac{1}{2} z left(cot ^2(x)+1right)+Oleft(z^2right)$$ which shows the limit and also how it is approached.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Claude Leibovici

                119k1157132




                119k1157132























                    1














                    This is not a solution, more of a question.



                    I get to



                    1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$



                    $y rightarrow infty : $



                    Known:



                    2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.



                    (Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)



                    Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?



                    Could not come up with a solution.



                    Thanks






                    share|cite|improve this answer























                    • The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                      – Paramanand Singh
                      yesterday












                    • Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                      – Peter Szilas
                      yesterday
















                    1














                    This is not a solution, more of a question.



                    I get to



                    1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$



                    $y rightarrow infty : $



                    Known:



                    2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.



                    (Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)



                    Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?



                    Could not come up with a solution.



                    Thanks






                    share|cite|improve this answer























                    • The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                      – Paramanand Singh
                      yesterday












                    • Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                      – Peter Szilas
                      yesterday














                    1












                    1








                    1






                    This is not a solution, more of a question.



                    I get to



                    1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$



                    $y rightarrow infty : $



                    Known:



                    2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.



                    (Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)



                    Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?



                    Could not come up with a solution.



                    Thanks






                    share|cite|improve this answer














                    This is not a solution, more of a question.



                    I get to



                    1)$log left ( cos (1/y) +frac{sin (1/y)}{1/y}frac{cot x}{y}right )^y.$



                    $y rightarrow infty : $



                    Known:



                    2) $lim_{y rightarrow infty} log (1+(cot x)/y)^y= cot x$.



                    (Recall $lim_{y rightarrow infty}(1+a/y)^y= e^a$, $a$ real.)



                    Is there a fairly straightforward way to get from 1)to 2) using limit manipulations ?



                    Could not come up with a solution.



                    Thanks







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited yesterday

























                    answered 2 days ago









                    Peter Szilas

                    10.8k2720




                    10.8k2720












                    • The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                      – Paramanand Singh
                      yesterday












                    • Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                      – Peter Szilas
                      yesterday


















                    • The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                      – Paramanand Singh
                      yesterday












                    • Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                      – Peter Szilas
                      yesterday
















                    The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                    – Paramanand Singh
                    yesterday






                    The way is not fairly straightforward but not that difficult either. Use the fact that if $f(y) >0$ and $y(f(y)-1) to 0$ as $ytoinfty $ then ${f(y)} ^yto 1$ as $yto infty$ (prove this!). Use $$f(y) =dfrac{cos(1/y)+sin(1/y)cot x} {1+dfrac{cot x} {y}} $$ and show that $y(f(y) - 1)to 0$ as $ytoinfty $.
                    – Paramanand Singh
                    yesterday














                    Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                    – Peter Szilas
                    yesterday




                    Paramand.Thanks a lot. I'll go through your suggestions. Was intrigued by [(cos 1/y)^y](1+[(cos (1/y))^{-1}]sin(1/y)(1/y)^{-1}[(cos x)/y])^y, By multiplication rule can consider lim (cos 1/y)^y×lim(1+A(y)(cos x)/y)^y, where lim A(y)=1, could not handle this second limit .
                    – Peter Szilas
                    yesterday


















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