Answer to Question #252511 in Real Analysis for dasuuu

Question

lim 𝑥→0+ (sin 𝑥) ^x

Answer 1

$\lim_{x\to\ 0^+} sin^x x= \lim_{x\to\ 0^+} e^{ln(sin^x )}$

Also,

$\lim_{x\to\ 0^+} e^{ln(sin^x x)}= \lim_{x\to\ 0^+} e^{xln(sinx)}$

On rearranging we get:

$\lim_{x\to\ 0^+} e^{xln(sin^x )}= e^{\lim_{x\to\ 0^+} \cfrac{ln(sinx)}{1/x}}$

Applying L’ hospital we get :

$e^{\lim_{x\to\ 0^+} \cfrac{ln(sinx)}{1/x}}= e^{\lim_{x\to\ 0^+} \cfrac{-x^2 cosx}{sinx}}$

on further rearranging we get:

$e^{\lim_{x\to\ 0^+} \cfrac{-x^2 cosx}{sinx}} = e^{\lim_{x\to\ 0^+} \cfrac{-x\ cosx}{\cfrac{sinx}{x}}}$

As ,

$\lim_{x\to\ 0^+} \cfrac{sinx}{x}=1$

Also

$\lim_{x\to\ 0^+} x\ cosx =0$

And hence ,

$e^{\lim_{x\to\ 0^+} \cfrac{-x\ cosx}{\cfrac{sinx}{x}}} = e^{0} =1$

hence,

$\lim_{x\to\ 0^+} sin^x x=1$