TITLE: Does $\lim f(x) = \frac{1}{\lim\frac{1}{f(x)}}$?
QUESTION [5 upvotes]: I was trying to find $\lim_{z\rightarrow{0}} \frac{z^2}{\cos(z)-1}$ and the solution guide just told me to do $\lim_{z\rightarrow{0}} \frac{\cos(z)-1}{z^2} = -\frac{1}{2}$  and then take the reciprocal of this, which is $-2$ as my answer.
Can we actually do that? Does this actually obey the limit laws?

REPLY [3 votes]: Here's another justification of it, just for fun.
In general, $$\lim_{x\to a} \frac{F(x)}{G(x)} = \frac{\displaystyle\lim_{x\to a} F(x)}{\displaystyle\lim_{x\to a} G(x)}$$
as long as both limits exist and $\displaystyle\lim_{x\to a} G(x) \ne 0$.
Applying that to your situation, we have:
$$ \lim_{x \to a} \frac1{f(x)} = \frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)} = \frac1{\displaystyle\lim_{x\to a} f(x)}$$
Therefore:
$$\frac1{\displaystyle\lim_{x\to a}\frac1{f(x)}} = \frac1{\left(\frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)}\right)} = \frac1{\left(\dfrac1{\displaystyle\lim_{x\to a} f(x)}\right)} = \lim_{x\to a} f(x)$$