Evaluating $\lim_{x \to \infty}\frac{(-1)^x}{x}$ with Euler's formula? 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.
 A: The number is bounded between -1 and 1 while the denominator approaches $\infty$. Hence, the limit must be zero (although it does not approach $0$ monotonically).
A: Let's just use an $\epsilon-\delta$ proof.  Choose $\epsilon \gt 0$.  Let $N = \lceil \frac 1\epsilon \rceil.$  Then $x \gt N \Rightarrow \lvert \frac{(-1)^x}{x} \rvert = \lvert \frac 1x \rvert \lt \epsilon$.
A: For  an arbitrary real number $x$, $(-1)^{x}$ is not well defined. 
For example if you let $x=1/24$ then you will have $24$ complex roots of $-1$
In order to make sense out of the question we need to modify it into a sequence  $$\lim_{n\to \infty} \frac {(-1)^n}{n}$$ in which case it is zero because its absolute value goes to $0$ 
A: You got the correct result, but I don't agree with your working.
Hint:
I would split this into two cases, $x=2k$, and $x=2k+1$. With this, you can better evaluate $(-1)^x$ to find the limit of both cases to be equal.
A: If we think of it as the limit of sequence
$$a_n=\frac{(-1)^n}{n}$$
so, in fact, it is equal to $0$, since
$$\left|\frac{(-1)^n}{n}\right|=\frac{1}{n}$$
and the last can be made as small as one wishes, for $n$ large enough.
