# Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$

I was trying to evaluate $$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$

I have tried taking natural logarithm first:

$$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\quad\text{(L'Hopital Rule)}\\ =\lim_{x\to0}\frac{\cos x}{x\sin x}-\frac{1}{x^2}\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)\\ =\lim_{x\to0}\frac{\cos x-1}{x^2}\quad\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)$$

and after this I eventually have the limit equaling $$-\frac{1}2$$, which means that the original limit should be $$\frac{1}{\sqrt{e}}$$.

However, I graphed $$f(x)=\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$ on Desmos, and it turned out that the limit is approximately $$0.7165313$$, or $$\frac{1}{\sqrt[3]{e}}$$.

Therefore I think there's something wrong in my approach, but I couldn't find it. Any suggestions?

• I buy your usage of LHR and the equation immediately after it, but jumping to $\lim_{x\to 0}\frac{\cos(x)-1}{x^2}$ is incorrect – Integrand Oct 19 '20 at 2:06
• @ Intergrand I see. So I can't apply $\lim_{x\to0}\frac{\sin x}x=1$ in that way? Any suggestions on what I should do instead? – Student1058 Oct 19 '20 at 2:08
• Get a common denominator and use LHR twice more; it works nice and good – Integrand Oct 19 '20 at 2:10
• Got it. Thank you! – Student1058 Oct 19 '20 at 2:17

Using the first terms of the Taylor series: $$\lim_{x\to 0}\left(\frac{\sin(x)}{x}\right)^{1/(1-\cos(x))}=\lim_{x\to 0}\left(1-\frac{x^2}{6}\right)^{1/(x^2/2)}$$Substitute $$z=2/x^2$$: $$=\lim_{z\to\infty} \left(1+\frac{-1/3}{z}\right)^{z}$$This is $$e^{-1/3}$$ by the limit definition of $$e^z$$, as shown here.

$$\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}=\lim_{x\to0}\frac{(x-\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots)-(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots)}{x^3}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$$

Let me try an approach without Taylor series, starting with your step

$$\lim_{x \to 0} \frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}=\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x}$$

We continue from here: \begin{align} \lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x} &\overset{\mathrm{H}}{=} \lim_{x \to 0} \ \frac{-x \sin x}{\sin^2x + 2x\sin x \cos x} \\ &= \lim_{x \to 0} \ \frac{-x}{\sin x + 2x \cos x} \overset{\mathrm{H}}{=} \lim_{x \to 0} \ \frac{-1}{3\cos x-2\sin x}=\frac{-1}{3} \end{align}

and the result follows from there.

Starting from this line of your answer: $$\ln L=\lim_{x\to0}\frac{\ln(\sin x /x)}{1-\cos x}$$ Apply l'Hospital's rule: $$\ln L=\lim_{x\to0}\frac{x(x\cos x -\sin x)}{x^2\sin^ 2x}=\lim_{x\to0}\frac{x^2(x\cos x -\sin x)}{x^3\sin^ 2x}$$ Now we have: $$\lim_{x\to0}\frac{(x\cos x -\sin x)}{x^3}$$ $$=\lim_{x\to0}\frac{-\sin x}{3x}=-\dfrac 13$$ Finally: $$\ln L=1 \times -\dfrac 13 \implies L=\dfrac 1 {\sqrt[3]e}$$