# 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.

• Observe that $|e^{i\theta}|=1$ for any $\theta$. – GReyes Jun 29 '19 at 0:31
• @GReyes: For any real $\theta$. – Ted Shifrin Jun 29 '19 at 0:46
• Right. Thanks for pointing out. – GReyes Jun 29 '19 at 1:11
• In the last, if $L=0$, then $\ln L=-\infty$ not $\infty$. – Hussain-Alqatari Sep 5 '19 at 4:40

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

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).

• Be careful. When $x$ is not an integer, $(-1)^x$ is a (multivalued) complex number, but any of their magnitudes will be $1$. – Ted Shifrin Jun 29 '19 at 0:45
• @Ted Shifrin Yes indeed. Thank you. I presumed the question pertained to real variables only. – mlchristians Jun 29 '19 at 0:47
• 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. – Ted Shifrin Jun 29 '19 at 0:51

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 $$0$$

• Modified into a sequnce? – user516076 Jun 29 '19 at 1:00
• Correct, that is what I had in mind. – Mohammad Riazi-Kermani Jun 29 '19 at 1:03

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.

• Why are you assuming that $x$ is an integer? – GReyes Jun 29 '19 at 0:36

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.

• Hmmm... so it use absolute convergence test? – user516076 Jun 29 '19 at 0:40
• 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 $|a_n-0|<\varepsilon$. – azif00 Jun 29 '19 at 0:51