# Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? [duplicate]

$$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$

My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox) \begin{align} & \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\ & \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\ & e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\ & e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\ & e^{-\infty} = 0\\ \end{align}

## Edit #1:

Continuing after the mistake of the $\tan(x)$:

\begin{align} & e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\ \end{align}

• $\frac{\cos x}{\sin x} = \cot x$, you wrote $\tan x$, hence got a wrong result. – Daniel Fischer Feb 2 '15 at 23:27
• @DanielFischer: I solve that problem but it still seems to be zero :/ – Dor Feb 3 '15 at 0:01
• Since $\cot$ has a pole at $0$, $\frac{\cot x}{x}$ is not not an indeterminate form if you plug in $x = 0$. It isn't one of the things you can apply L'Hospital's rule to. You can't separate the two summands, write the exponent as $\frac{x\cot x - 1}{x^2}$. But better, use the Taylor expansions rather than L'Hospital's rule, that's easier and less error-prone. – Daniel Fischer Feb 3 '15 at 0:03
• Just a suggestion on formatting: consider using $\exp(f(x))$ in place of $e^{f(x)}$. The latter can make rational exponents tough to read. – user170231 Feb 3 '15 at 0:36
• @user170231: Thx didn't thought of that! Though, BTW, it is possible to zoom in a MathJax code (at least in Firefox). – Dor Feb 3 '15 at 9:02

Take the logarithm of your limit then use L'Hopital and Taylor series. Let the expression in the limit be $L$:

\begin{align} \ln L&=\ln\left[ \left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \right] \\[2ex] &= \frac{\ln\left(\frac{\sin(x)}{x}\right)}{x^2} \\[2ex] &\to\frac{\frac{x}{\sin x}\cdot \frac{x\cos x-\sin x}{x^2}}{2x}\qquad \text{(L'Hopital)}\\[2ex] &= \frac{x\cos x-\sin x}{2x^2\sin x} \\[2ex] &\to\frac{x(1-\frac{x^2}2)-(x-\frac{x^3}6)}{2x^2(x)}\qquad \text{(Taylor series)} \\[2ex] &= \frac{-\frac 13x^3}{2x^3} \\[2ex] &= -\frac 16 \end{align}

Therefore $L\to e^{-1/6}$. (I left a few minor gaps for you to fill in.)

If you don't like Taylor series, we can finish the whole thing with L'Hopital, though this way takes more steps.

\begin{align} \ln L&\to\frac{x\cos x-\sin x}{2x^2\sin x} \qquad\text{(from above, before Taylor series)}\\[2ex] &\to\frac{x\cdot -\sin x +\cos x-\cos x}{2x^2\cos x + 4x\sin x} \qquad\text{(L'Hopital)}\\[2ex] &= \frac{-\sin x}{2x\cos x + 4\sin x} \\[2ex] &\to\frac{-\cos x}{2x\cdot -\sin x+2\cos x+4\cos x} \qquad\text{(L'Hopital)}\\[2ex] &= \frac{-\cos x}{-2x\sin x+6\cos x} \\[2ex] &\to\frac{-1}{0+6} \\[2ex] &= -\frac 16 \end{align}

• That solution really seems good but I'm curious as to how we solve that limit without Taylor..? – Dor Feb 3 '15 at 9:04
• @Dor: See my edited answer for a solution with Taylor series. Does this meet your needs? – Rory Daulton Feb 3 '15 at 12:57
• Instead of doing the “hard” derivative of $\log((\sin x)/x)$, one can more simply write it as $\log|\sin x|-\log|x|$ and the derivative is $\cot x-1/x=(x\cot x-1)/x$ – egreg Feb 3 '15 at 13:25
• @RoryDaulton: Indeed thanks! :) – Dor Feb 3 '15 at 18:24
• @egreg: You are right, but I still prefer the way I showed. Your way leads to $\cot$ and $\csc$ functions, and I avoid them whenever possible--my personal preference. And if you don't simplify the $\cot$ away, the later differentiations become more difficult. – Rory Daulton Feb 3 '15 at 19:29

Using Taylor series:

$$\frac{\sin(x)}{x}=1- \frac{1}{3!}x^2+o(x^2).$$
So using this formula we get: $$\ln \left( \frac{\sin(x)}{x}\right)=\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right).$$
Now we can use the Taylor expansion for the $\ln$ function so we get: $$\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right)= - \frac{1}{3!}x^2+o(x^2).$$ Now divide all by $x^2$ and compute the limit while $x$ goes to zero: $$\frac{- \frac{1}{3!}x^2+o(x^2)}{x^2}$$ as $x\rightarrow 0$, we obtain that $$\lim_{x \rightarrow 0} \ln \left( \frac{\sin(x)}{x}\right)= -\frac{1}{6}.$$ At the end $$\lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x} \right)^{\frac{1}{x^2}}= e^{\lim_{x \rightarrow 0}\frac{\ln\frac{\sin(x)}{x}}{x^2}}= e^{-\frac{1}{6}}.$$

:( I have not enough reputation to even add a comment...

Yes, the cot<-->tan problem.

Comment: it should be equal to $e^{-\frac{1}{6}}$, if the cot<-->tan was not there.

• Wolfra Alpha also gets $1/(\sqrt[6]{e})$. – Jeppe Stig Nielsen Feb 2 '15 at 23:50
• I solved that problem but it still seems to be 0... though Wolfram Alpha gets it otherwise. Where did I go wrong..? – Dor Feb 3 '15 at 0:00

\begin{align*} \left(\frac{\sin x}x\right)^{\frac1{x^2}} &=\left[1+\left(\frac{\sin x}x-1\right)\right]^{\frac1{x^2}}\\ &=\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{\frac1{x^2}}\\ &=\left[\underbrace{\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{[x/(\sin x-x)]}}_{=:A(x)}\right]^{\frac{\sin x-x}{x^3}} \end{align*}

Next observe that

$\bullet\sin x=x-\frac{x^3}{6}+o(x^3) \Longrightarrow\frac{\sin x-x}{x^3}\stackrel{x\to0}{\longrightarrow}-\frac16$;

$\bullet\lim_{\xi\to\infty}\left(1+\frac1\xi\right)^{\xi}=e \Longrightarrow A(x)\stackrel{x\to0}{\longrightarrow} e \;\;(\mbox{in fact} \;\;x/(\sin x-x)\stackrel{x\to0}{\longrightarrow}\infty)$

thus we immediately get $$\lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-\frac16}$$

• That's a really nice method..! – Dor Feb 3 '15 at 18:27
• Many thanks Dor! – Joe Feb 3 '15 at 22:24