How to solve this limit?
$$\lim_{x \to \infty}\frac{(-1)^x}{x}$$
Wolfram Alpha said that the answer is $0$.
But I think that it has no limit. What I tried is
$\begin{align} \lim\limits_{x \to \infty}\dfrac{(-1)^x}{x} = \lim\limits_{x \to \infty}\dfrac{e^{i\pi x}}{x} \end{align}$
I have no idea how to get $0$ from that, because it looks like undefined $$(\frac{\infty}{\infty})$$
Or, I have another perspective like this:
$\begin{align}\text{Let}\, \dfrac{(-1)^x}{x}&=L\\ \ln L &= \ln \left(\dfrac{e^{i\pi x}}{x}\right)\\ \lim\limits_{x \to \infty}\ln L &= \lim\limits_{x \to \infty}\ln \left(\dfrac{e^{i\pi x}}{x}\right)\\ &=\lim\limits_{x \to \infty}i\pi x - \ln x\\ &=\infty - 0\\ &=\infty \end{align}$
Thus
$\begin{align} \lim\limits_{x \to \infty}\ln L&=\infty\\ L=0 \end{align}$
I don't know if it's right or not. Please do a correction. Thanks.