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.
Best Answer
Let's just use an $\epsilon-\delta$ proof. Choose $\epsilon \gt 0$. Let $N = \lceil \frac 1\epsilon \rceil.$ Then $x \gt N \Rightarrow \lvert \frac{(-1)^x}{x} \rvert = \lvert \frac 1x \rvert \lt \epsilon$.