Question: We were asked to find the limit of $(\frac{1}{x})^{\sin x}$. The following picture is the solution shown by our mentor. But I have a different approach.

![](https://i.stack.imgur.com/O7xie.jpg)
$$\lim\_{x \to 0} \left(\frac{1}{x}\right)^{\sin x}=e^{\lim\limits\_{x \to 0} ((\frac{1}{x}-1)\sin x)}=e^{\lim\limits\_{x \to 0} (1-x)\frac{\sin x}{x}}=e.$$
Both of the calculations look fine to me, where did I go wrong then?

Answer: 
By definition, if the domain of $\gamma$ is $[a,b]$, then$$\int\_\gamma\frac1{z^2}\,\mathrm dz=\int\_a^b\frac{\gamma'(t)}{\gamma^2(t)}\,\mathrm dt$$and it seems to me that you are not taking that $\gamma'(t)$ into account.

To be more concrete, take $\gamma\colon[0,\pi]\longrightarrow\Bbb C$ defined by $\gamma(t)=e^{it}$ and take $\eta\colon[\pi,2\pi]\longrightarrow\Bbb C$ also defined by $\eta(t)=e^{it}$. Then\begin{align}\int\_\eta\frac1{z^2}\,\mathrm dz&=\int\_\pi^{2\pi}\frac{ie^{it}}{e^{2it}}\,\mathrm dt\\&=\int\_0^\pi\frac{ie^{i(t+\pi)}}{e^{2i(t+\pi)}}\,\mathrm dt\\&=-\int\_0^\pi\frac{ie^{it}}{e^{2it}}\,\mathrm dt,\end{align}since $e^{\pi i}=-1$ and $e^{2\pi i}=1$. But the final integral is just $\int\_\gamma\frac1{z^2}\,\mathrm dz$.

The same argument applies if we have any even function instead of $z^2$.
After the substitution $z=e^{i\theta} \rightarrow dz=ie^{i\theta}d\theta$, you're left with $I=i\int\_0^\pi e^{-i\theta}d\theta=2$