Find the value of $\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$. [duplicate]

I am stuck with the following problem :

Find the value of $$\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$$

My try : Let $$p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}{\frac {1}{x^2}}\log\left(\frac {\sin x}{x}\right)$$...After applying l'hospitals rule few times I get

$$\log p= \lim_{x \to 0}\frac{1-\sec^2x}{4x \tan x+2x^2\sec^2x}$$.. we can again apply l'hospitals rule ,but the calculations get bigger and bigger..Is there any other easier way around or I am missing something?

marked as duplicate by Namaste, Leucippus, Claude Leibovici limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 29 '17 at 6:05

From the Taylor expansion of $\sin x$ around $0$, namely $\sin u = u -\frac{u^3}{6} + o(u^3)$; and that of $\ln(1+u)$, specifically $\ln(1+u)=u+o(u),$ we get \begin{align} \left( \frac{\sin x}{x}\right)^{1/x^2} &= \left(1-\frac{x^2}{6} + o(x^2)\right)^{1/x^2} = e^{\frac{1}{x^2}\ln\left(1-\frac{x^2}{6} + o(x^2)\right) } = e^{\frac{1}{x^2}\left(-\frac{x^2}{6} + o(x^2)\right) }\\ &= e^{-\frac{1}{6} + o(1)} \xrightarrow[x\to0]{}\boxed{e^{-\frac{1}{6}}} \end{align}

• awesome..got it – learner Aug 28 '17 at 19:29
• Glad that helped! – Clement C. Aug 28 '17 at 19:29

Applying L'Hospital's rule rwice we will get

$$\lim _{ x\to 0 } \left( \frac { \sin x }{ x } \right) ^{ \frac { 1 }{ x^{ 2 } } }=\lim _{ x\to 0 } \left[ { \left( 1+\frac { \sin x-x }{ x } \right) }^{ \frac { x }{ \sin { x-x } } } \right] ^{ \frac { \sin { x-x } }{ x^{ 3 } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x-x } }{ x^{ 3 } } } }=\\ \overset { L'hospital's }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3x^{ 2 } } } }\overset { \quad L'hospital's }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x } }{ 6x } } }=\color{red}{{ e }^{ -\frac { 1 }{ 6 } }}$$

You can do this with L'Hopital, if you massage the expression you get from the first round a bit:

$${f(x)\over g(x)}={\log\sin x-\log x\over x^2}\implies{f'(x)\over g'(x)}={{\cos x\over\sin x}-{1\over x}\over2x}={x\cos x-\sin x\over2x^2\sin x}$$

At worst, another three rounds of L'Hopital will leave something nonzero in the denominator. You can speed things up by noting that

$${(x\cos x-\sin x)'\over(2x^2\sin x)'}={-x\sin x\over4x\sin x+2x^2\cos x}={-1\over4+2{x\over\sin x}\cos x}\to{-1\over4+2\cdot1\cdot1}={-1\over6}$$