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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.

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    $\begingroup$ Observe that $|e^{i\theta}|=1$ for any $\theta$. $\endgroup$
    – GReyes
    Jun 29, 2019 at 0:31
  • $\begingroup$ @GReyes: For any real $\theta$. $\endgroup$
    – Ted Shifrin
    Jun 29, 2019 at 0:46
  • $\begingroup$ Right. Thanks for pointing out. $\endgroup$
    – GReyes
    Jun 29, 2019 at 1:11
  • $\begingroup$ In the last, if $L=0$, then $\ln L=-\infty$ not $\infty$. $\endgroup$
    – Hussain-Alqatari
    Sep 5, 2019 at 4:40

5 Answers 5

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

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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).

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  • $\begingroup$ Be careful. When $x$ is not an integer, $(-1)^x$ is a (multivalued) complex number, but any of their magnitudes will be $1$. $\endgroup$
    – Ted Shifrin
    Jun 29, 2019 at 0:45
  • $\begingroup$ @Ted Shifrin Yes indeed. Thank you. I presumed the question pertained to real variables only. $\endgroup$
    – mlchristians
    Jun 29, 2019 at 0:47
  • $\begingroup$ Yes, but remember that $a^x = e^{x\log a}$, and it's only because $a=-1$ that all the values of $\log a$ are purely imaginary. $\endgroup$
    – Ted Shifrin
    Jun 29, 2019 at 0:51
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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 $0$

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  • $\begingroup$ Modified into a sequnce? $\endgroup$
    – user516076
    Jun 29, 2019 at 1:00
  • $\begingroup$ Correct, that is what I had in mind. $\endgroup$
    – Mohammad Riazi-Kermani
    Jun 29, 2019 at 1:03
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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.

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    $\begingroup$ Why are you assuming that $x$ is an integer? $\endgroup$
    – GReyes
    Jun 29, 2019 at 0:36
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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.

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  • $\begingroup$ Hmmm... so it use absolute convergence test? $\endgroup$
    – user516076
    Jun 29, 2019 at 0:40
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    $\begingroup$ No. It use the definition of the limit of a sequence, that is, for every $\varepsilon>0$ there exists a natural number $N$ such that, for all $n\geq N$ we have $|a_n-0|<\varepsilon$. $\endgroup$
    – azif00
    Jun 29, 2019 at 0:51

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