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My calculus book states that $\lim\limits_{x\rightarrow \infty} (-1^x\frac{1}{2^{x-1}})=0$. However, when verifying the calculation by Wolfram, I get a result that the limit does not exist.

What is the correct answer?

David G. Stork
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Martin
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    $\lim\limits_{x\to\infty}(-1^x\frac{1}{2^{x-1}})=\lim\limits_{x\to\infty}2\cdot (-\frac{1}{2})^x\to 0$. It should be well known that $\lim\limits_{x\to\infty}r^x$ where $|r|<1$ always goes to zero. – JMoravitz Apr 19 '20 at 18:51
  • See here for the proof that $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$. – Axion004 Apr 19 '20 at 19:05
  • Notice that for real numbers $(-1)^x$ is defined only when $x = p/q$, where $p,q \in \mathbb Z$ and $q$ is odd. – Sam Apr 20 '20 at 00:54

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