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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?

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  • $\begingroup$ 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 $\endgroup$ – Integrand Oct 19 '20 at 2:06
  • $\begingroup$ @ 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? $\endgroup$ – Student1058 Oct 19 '20 at 2:08
  • $\begingroup$ Get a common denominator and use LHR twice more; it works nice and good $\endgroup$ – Integrand Oct 19 '20 at 2:10
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    $\begingroup$ Got it. Thank you! $\endgroup$ – Student1058 Oct 19 '20 at 2:17
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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.

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

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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.

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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: $$\ln L=1 \times -\dfrac 13 \implies L=\dfrac 1 {\sqrt[3]e}$$

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