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:
$$\frac1{\displaystyle\lim_{x\to a}\frac1{f(x)}} = \frac1{\left(\frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)}\right)} = \frac1{\left(\dfrac1{\displaystyle\lim_{x\to a} f(x)}\right)} = \lim_{x\to a} f(x)$$