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As the title states, why does$\lim\limits_{x\to\infty}\dfrac{(-1)^{x+1}}{x}$? It seems like this would produce the indeterminate form of $\frac{\infty}{\infty}$ at which point I would use L'Hospital's rule but I don't even know how to take the derivative of $(-1)^{x+1}$.

Could anyone show me why $\lim\limits_{x\to\infty}\dfrac{(-1)^{x+1}}{x}$ converges to 0? (According to Wolfram Alpha).

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$(-1)^{x+1}$ is always $\pm 1$, it doesn't go to $\infty$. – Alex Becker Apr 4 '14 at 6:00
L'Hospital obsession at its best. – Did Apr 4 '14 at 6:01
We should really be using $n$ instead of $x$: the limit needs to go through integer values. Otherwise we'd have something like $(-1)^{1/2}$. – user134824 Apr 4 '14 at 6:02
Who can say $(-1)^{x+1}=\pm 1$? How we can say $(-1)^{\sqrt{2}}=\pm 1$? – David Apr 4 '14 at 6:14
Picky remark: it's not the limit that converges to zero, but the expression which we're taking the limit of. The limit itself equals zero. (We say "I'm going to Spain" or "my destination is Spain", not "my destination is going to Spain".) – Hans Lundmark Apr 4 '14 at 7:10
up vote 4 down vote accepted

We have

$$0\le\left|\dfrac{(-1)^{x+1}}{x}\right|=\dfrac{1}{x}\xrightarrow {x\to\infty} 0$$

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When you don't know the value of $(-1)^\sqrt{2}$ how you can use $|.|$ for it. I think $|.|:\Bbb{R}\to\Bbb{R}^+$. – David Apr 4 '14 at 6:18

First note that $$ \lim_{x\to\infty} \frac{1}{x}=0\quad \mbox{and}\quad \lim_{x\to\infty} \frac{-1}{x}=0 $$ Therefore $$ \lim_{x\to\infty} \frac{(-1)^{x+1}}{x}=0 $$

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