Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.
 A: With l'Hospital directly:
\begin{align*}
\lim_{x\to 0}\frac{\log\frac{\sin x}x}{1-\cos x} & \stackrel{\text{l'H}}=\lim_{x\to 0}\frac{\frac 1{\sin x}\cdot\frac{x\cos x-\sin x}x}{\sin x} \\
& =\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin^2x}\\
& \stackrel{\text{l'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}\\&=-\lim_{x\to 0}\frac1{\frac{\sin x}x+2\cos x}\\
&=-\frac13
\end{align*}
A: One may recall that, as $x \to 0$, we have
$$
\cos x =1-\frac {x^2}{2}+\mathcal{O}(x^3), \quad
\sin x =x-\frac{x^3}{6}+\mathcal{O}(x^4), \quad
\ln (1+x)=x+\mathcal{O}(x^2),
$$ giving
$$
\begin{align}
\frac{1}{1-\cos x}&=\frac 2{x^2}+\mathcal{O}(1)\\\\
\ln \left(\frac{\sin x}x\right)& =-\frac{x^2}{6}+\mathcal{O}(x^3).
\end{align}
$$ Then, as $x \to 0$,
$$
\begin{align}
\left(\frac{\sin x}x\right)^{\small \dfrac{1}{1-\cos x}}&=e^{\Large \frac{1}{1-\cos x}\ln \left(\frac{\sin x}x\right)}\\\\
& =e^{\large \left(\frac 2{x^2}+\mathcal{O}(1)\right)\left(-\frac{x^2}{6}+\mathcal{O}(x^3)\right)}\\\\
& =e^{\Large -\frac 13+\mathcal{O}(x)}
\end{align}
$$ and the desired limit is $\displaystyle e^{-\large \frac 13}$.
A: Let us start with $$A=\left(\frac{\sin (x)}{x}\right)^{\frac{1}{1-\cos (x)}}$$ Take logarithms; so $$\log(A)=\frac{1}{1-\cos (x)}\log\left(\frac{\sin (x)}{x}\right)$$ Now, use Taylor series $$\sin(x)\approx x-\frac{x^3}{6}$$ $$\frac{\sin (x)}{x}\approx 1-\frac{x^2}{6}$$ Now consider that,for small $y$, $\log(1+y)\approx y-\frac{y^2}{2}$ So $$\log\left(\frac{\sin (x)}{x}\right)\approx -\frac{x^2}{6}-\frac{x^4}{180}$$ Also $$\cos(x)\approx 1-\frac{x^2}{2}+\frac{x^4}{24}$$ Using all of the above, we then have $$\log(A)\approx \frac{-\frac{x^2}{6}-\frac{x^4}{180}}{\frac{x^2}{2}-\frac{x^4}{24}}= \frac{-\frac{1}{6}-\frac{x^2}{180}}{\frac{1}{2}-\frac{x^2}{24}} \approx -\frac{1}{3}-\frac{7 x^2}{180}$$ which shows the limit of $\log(A)$ and how it is approached.
I am sure that you can take from here.
A: If $L$ is the desired limit then
\begin{aligned}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/(1 - \cos x)}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{\sin x}{x}\right)^{1/(1 - \cos x)}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{1 - \cos x}\cdot\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{1 + \cos x}{\sin^{2} x}\cdot\log\left(1 + \frac{\sin x}{x} - 1\right)\notag\\
&= \lim_{x \to 0}(1 + \cos x)\cdot\frac{x^{2}}{\sin^{2} x}\dfrac{\dfrac{\sin x}{x} - 1}{x^{2}}\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= \lim_{x \to 0}(1 + 1)\cdot 1\cdot\frac{\sin x - x}{x^{3}}\cdot 1\notag\\
&= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (by L'Hospital Rule)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{x^{2}(\cos x + 1)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\left(-\frac{\sin^{2}x}{x^{2}}\right)\frac{1}{1 + \cos x}\notag\\
&= -\frac{2}{3}\cdot\frac{1}{1 + 1} = -\frac{1}{3}\notag
\end{aligned}
It follows that $L = e^{-1/3}$.
A: The first thing that you see is that this is a limit of the form $1^{\infty}$. Generally these limits are like $\lim_{x \to 0} f(x)^{g(x)}$ where $\lim_{x\to 0}f(x)=1\pm\epsilon$ and $\lim_{x\to 0}g(x)=\infty$. the standard way to solve these types of limits is in the following way:-$$\begin{align}\lim_{x\to 0}f(x)^{g(x)}&=\lim_{x\to 0}e^{g(x)(\ln[f(x)])}\\&=\lim_{x\to 0}e^{g(x)(\ln[1\pm\epsilon])}\\&=\lim_{x\to 0}e^{g(x)(\pm\epsilon)}\\&=\lim_{x\to 0}e^{g(x)(f(x)-1)}\end{align}$$ You have to proceed from here. For this problem it is reduced to $$\lim_{x\to 0}e^{\frac 1{1-\cos x}(\frac {\sin x}x-1)}=\lim_{x\to 0}e^{\frac {\sin x-x}{x(1-\cos x)}}=\lim_{x\to 0}e^{\frac {x-\frac{x^3}6-x}{x(1-1+\frac {x^2}2)}}=\lim_{x\to 0}e^{\frac {-\frac{x^3}6}{\frac {x^3}2}}=e^{\frac{\frac {-1}{6}}{\frac 12}}=e^{\frac {-1}3}$$In the $3rd$ step I have used the expansion of $\sin x$ and $\cos x$ as $x\to 0$.
