Does $\lim f(x) = \frac{1}{\lim\frac{1}{f(x)}}$? 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?
 A: This is the justification. Suppose $g$ is continuous at $L$, and suppose $\lim\limits_{x\to a}f(x)=L$. Then
$$\lim\limits_{x\to a}g(f(x)) = g(\lim\limits_{x\to a}f(x)) = g(L)$$
The first equality is from the continuity of $g$ at $L$, and the second is the existence of the original limit.
One needs to be a little more careful in the cases $x\to\pm\infty$ since the domain of $g$ typically doesn't include $\infty$.
In your case, $g(x)=1/x$, and your statement can be rearranged to be
$$g(\lim f(x)) = \lim g(f(x))$$ by inverting both sides.
A: 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:
$$\frac1{\displaystyle\lim_{x\to a}\frac1{f(x)}} = \frac1{\left(\frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)}\right)} = \frac1{\left(\dfrac1{\displaystyle\lim_{x\to a} f(x)}\right)} = \lim_{x\to a} f(x)$$
