Can we move the limit out of the reciprocal of the reciprocal? Say we have a function f(x) from R to R defined at c and it is continuous at the point so $\lim\limits_{x \to c}{f{(x)}} = {f{(c)}}$ Are we allowed to say that $\frac{1}{\frac{1}{\lim\limits_{x \to c}{f{(x)}}}} = \lim\limits_{x \to c}\frac{1}{\frac{1}{{f{(x)}}}}$
There are some specific proofs that would be much easier with this trick and I was wondering when we are allowed to do this. Is the only condition that $f(c) \neq 0$ enough?
 A: If $\lim_{x\rightarrow c} f(x)=f(c)\neq 0$ then
$$
\frac{1}{\frac{1}{\lim_{x\to c} f(x)}}=\lim_{x\to c} f(x)=\lim_{x\to c}\frac{1}{\frac{1}{f(x)}}.
$$
A: If $f(c) \neq 0$ and $f$ is continuous at $c$ then it means that,
$$\forall \varepsilon >0, \exists \eta>0, \forall x \in \mathbb{R}, \,|x-c|\leq \eta \implies|f(x)-f(c)|\leq\varepsilon$$
In particular, let's assume WLOG that $f(c)>0$ then by taking $\varepsilon_0 = f(c)/2>0$ there exists $\eta_0>0$ such that for all $x \in \mathbb{R}$ verifying $|x-c|\leq \eta_0$ we have $|f(x)-f(c)|\leq f(c)/2$.
Hence,
$$\forall x \in [c-\eta_0,c+\eta_0], -f(c)/2 \leq f(x)-f(c) \leq f(c)/2$$
Therefore,
$$\forall x \in [c-\eta_0,c+\eta_0],0<f(c)/2\leq f(x) \leq 3f(c)/2$$
In particular,
$$\forall x \in [c-\eta_0,c+\eta_0], f(x) \neq 0$$
So you can find a neighbourhood of $c$ where $f$ is always nonzero, let's denote it by $V = [c-\eta_0,c+\eta_0]$.

Extra proof: As subrosar mentioned this part is not necessary to prove what you ask, it is just a proof that $\displaystyle \lim_{x \to c} \dfrac{1}{f(x)} =  \dfrac{1}{f(c)}$ as it might be an interesting result to keep in mind.
Let $\hat{\varepsilon}>0$ be fixed then for all $x \in V$ we have,
$$ \left|\frac{1}{f(c)}-\frac{1}{f(x)} \right| \leq \hat{\varepsilon}  \iff |f(x)-f(c)|\leq \hat{\varepsilon} |f(x)f(c)|$$
Thus by taking $\varepsilon_1 = \hat{\varepsilon}f(c)^2/2$ there exists $\eta_1>0$ such that,
$$\forall x \in V  \cap [c-\eta_1,c+\eta_1], |f(x)-f(c)|\leq \varepsilon_1 \leq \hat{\varepsilon} |f(x)f(c)| $$
Therefore it proves, if we call $\hat{\eta} = \min(\eta_0,\eta_1)$ that,
$$\forall x \in \mathbb{R}, |x-c|\leq \hat{\eta} \implies \left|\frac{1}{f(x)}-\frac{1}{f(c)}\right| \leq \hat{\varepsilon} $$
Hence,
$$\lim_{x \to c} \dfrac{1}{f(x)}=\frac{1}{f(c)}$$

In any case as we have, $f(c) \neq 0$ then,
$$ \dfrac{1}{\lim_{x \to c} f(x)} = \dfrac{1}{f(c)} \neq 0$$
Hence,
$$ \dfrac{1}{\dfrac{1}{\lim_{x \to c} f(x)}} = \dfrac{1}{1/f(c)} = f(c) = \lim_{x \to c} f(x) = \lim_{x \to c} \dfrac{1}{1/f(x)} $$
And $\displaystyle \lim_{x \to c} \dfrac{1}{1/f(x)}$ makes sense because $f$ does not take value $0$ on $V$ a neighbourhood of $c$.
