$\displaystyle\lim_{x \to a}(f(x))^n = (\lim_{x \to a}f(x))^n$ I want to prove $\displaystyle\lim_{x \to a}(f(x))^n = (\lim_{x \to a}f(x))^n$ and then use this to prove that $\displaystyle\lim_{x \to a}\sqrt{f(x)} = \sqrt{\lim_{x \to a}f(x)}$.
For the first one, I have seen induction (well sort of) applied to this if we assume the limit of the product is the product of the limits is already proven. It goes something like:
Base Case: $n=2$
Then $\displaystyle\lim_{x \to a} = (f(x))^2 = \lim_{x \to a}f(x)\cdot \lim_{x \to a}f(x)$
Inductive Case: Suppose $n = k$.
Then $\displaystyle\lim_{x \to a}(f(x))^k = \lim_{x \to a}(f(x)f(x)^{k-1})$
Ya so it gets sort of confusing to me for the $n = k$ case assuming this is the right way to go about this. Intuitively, it makes sense. Your just multiplying $f(x)$ $n$ times, but I am having trouble with this proof if correct. And I am not sure how to make the connection with the square root one.  
 A: In general for a continuous function $g$ $$\lim \limits_{x\rightarrow a}g(f(x))= g\Big(\lim \limits_{x\rightarrow a}f(x)\Big)$$ 
 The function $y=\sqrt x$ for is continuous. I think you can take it from here.
A: You are definitely on the right track for proving $\lim_{x \rightarrow a}(f(x))^n = ( \lim_{x \rightarrow a}(f(x)))^n$.  When doing an inductive proof you start by proving the base case, like you did.  Then you invoke the inductive hypothesis, which assumes the proof is true for the case $n = k-1$, and then you prove it is true for the case $n=k$.
So assuming the inductive hypothesis we have
$$
\lim_{x \rightarrow a}(f(x))^{k-1} = ( \lim_{x \rightarrow a}f(x))^{k-1}.
$$
Taking what you have and assuming the limit of products is the product of limits, we can show,
\begin{align}
\lim_{x \rightarrow a} (f(x))^k &= \lim_{x \rightarrow a}(f(x)(f(x))^{k-1}) \\
&= \lim_{x \rightarrow a} f(x) \lim_{x \rightarrow a}(f(x))^{k-1} \\
&= \lim_{x \rightarrow a} f(x) (\lim_{x \rightarrow a}f(x))^{k-1} \\
&= (\lim_{x \rightarrow a}f(x))^{k}
\end{align}
Done!
Now to show the square root commutes with the limit requires more work, since the previous proof only applies to integers $n$.  But we can use the previous proof to help us.  Starting with $\lim_{x \rightarrow a}f(x)$,
$$
\lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a}(\sqrt{f(x)})^2 = (\lim_{x \rightarrow a}\sqrt{f(x)} \;)^2
$$
where the last equality is obtained by the previous theorem.  Taking the square root of both sides gives your result.
A: $\lim_{x\to a} (f(x))^n=\lim_{x\to a} \{f(x)f(x)f(x)f(x) \dots f(x)\}$(n times)
separate limits
$\lim_{x\to a} (f(x))^n=\lim_{x\to a} f(x)\lim_{x\to a} f(x)\lim_{x\to a} f(x) \dots \lim_{x\to a} f(x)$(n times)
$\lim_{x\to a} (f(x))^n=(\lim_{x\to a} f(x))^n $
