Determining the limit of the function composition of a known limit For example, given $$\lim_{n \to \infty}\frac{n^2}{c+n^2} = 1,$$ are we allowed to "square both sides" $$\lim_{n \to \infty}\left(\frac{n^2}{c+n^2}\right)^2 = 1^2,$$ basically finding finding the limit of the composition $g f,$ where $g = x^2$ and $f = \frac{n^2}{x+n^2}.$
 A: Partial answer:
Basically, we'll try to find
$$\lim_{n \to \infty}\left(\frac{n^2}{n^2+c}\right)^x$$ were $n \in \Bbb Z^+$.
By the binomial theorem, this is just:
$$\lim_{n \to \infty}\frac{n^{2x}}{\sum_{k=0}^x\binom{x}{k}n^{2k}c^{x-k}}$$
We can see that the leading coefficient of the polynomial in the denominator is $n^{2x}$, and both the numerator and denominator are of the same degree. So, I'd like to quote Mark Ryan in 'Calculus for Dummies':

To impress your friends, point your index finger upward, raise one
eyebrow, and say in a professonial tone, “In a rational function
where the numerator and denominator are of equal degrees, the limit of
the function as $x$ approaches infinity or negative infinity equals
the quotient of the coefficients of the leading terms. A horizontal
asymptote occurs at this same value.”

(Emphasis mine). Since $n^{2x}/n^{2x} = 1$, we have:
$$\lim_{n \to \infty}\left(\frac{n^2}{n^2+c}\right)^x = 1$$
A: 
For example, given $$\lim_{n \to \infty}\frac{n^2}{c+n^2} = 1,$$ are we allowed to "square both sides" $$\lim_{n \to \infty}\left(\frac{n^2}{c+n^2}\right)^2 = 1^2,$$ basically finding finding the limit of the composition $g f,$ where $g = x^2$ and $f = \frac{n^2}{x+n^2}.$

Yes this is valid, because $x^2$ is continuous at $1.$
In general, determining the limit, as $x$ tends to $c$ or $\infty$ or $-\infty,$ of a composed function $gf,$ where $f$ has limit $l,$ in this way is valid provided that either of these conditions is satisfied:

*

*$g$ is continuous at $l$

*in every punctured neighbourhood of $c,\,$ $f(x) \neq l$.

A: The function of the limit is not guaranteed to be the limit of the function generally speaking. Pragmatically, if the right and left limits approach the same finite value in the function's domain, then the composition of the function's limit exists.
Concisely, $\lim_{x\to n} f(g(x)) = f(c)$ while $\lim_{x\to ^-n} g(x) = c$ and $\lim_{x\to ^+n} g(x) = c$ where $c \in R$ or $-\infty < c < \infty$.
As Elliot Yu kindly pointed out, it is only so if you can prove that the composition is continuous at the limit, and the limit exists. This should be trivial because it is the very definition of a continuous function that whenever $x_n \rightarrow x$ in the domain of a function $f$, then the limit $f(a_n)$ exists, and this limit is even $f(a)$.
Or generally speaking, if you are familiar with topology:

Let $f: X\rightarrow Y$ be a continuous function between topological spaces $X$ and $Y$. By definition, for each open set $V$ in $Y$, the preimage $f^{-1}(V)$ is open in $X$.
Now suppose $a_n \rightarrow a$ is a sequence with limit $a$ in $X$. Then $f(a_n)$ is a sequence in $Y$, and $f(a)$ is some point.
Choose a neighborhood $V$ of $f(a)$. Then $f^{-1}(V)$ is an open set (by continuity of $f$) which in particular contains $a$, and therefore $f^{-1}(V)$ is a neighborhood of $a$. By the convergence of $a_n$ to $a$, there exists an $N$ such that for $n > N$, we have $a_n \in f^{-1}(V)$.
Then applying $f$ to both sides gives that, for the same $N$, for each $n > N$ we have $f(a_n) \in V$. Originally $V$ was an arbitrary neighborhood of $f(a)$, so $f(a_n) \rightarrow f(a)$.

In other words, I originally said $\lim_{x\to n} g(f(x))$ is not always $g(\lim_{x\to n} f(x))$. But, they are equivalent in many cases.
Still, I was wrong to say that

even for convergent limits, let $\lim_{x\to n} f(x)=c$, then $\lim_{x\to n} g(f(x))$ is not necessarily $g(c)$.

This was wrong in my case because, as in my counter-example below, the limits do not converge. Infinity is not part of the real numbers, so there does not exist any $g(c)$ where $c=\infty$.
For example, let $f(x)=x^2$ and $g(x)=1/x$. Then, $\lim_{x\to0} f(g(x)) = \infty$ while $\lim_{x\to ^-0} g(x) = -\infty$ and $\lim_{x\to ^+0} g(x) = \infty$, so the limit does not exist.
There are limit laws for operations in and among limits such as
$$\lim_{x\to n} f(x)+g(x) = \lim_{x\to n} f(x) + \lim_{x\to n} g(x)$$
These can make more sense when considering the $\epsilon$ $\delta$ side of limits that only involve inequalities, which another answer might have more authority to speak on.
At risk of too many quotes:

Suppose $f: R \rightarrow R$ is a function defined on the real line, and there are two real numbers $p$ and $L$. One would say that "the limit of $f$, as $x$ approaches $p$, is $L$" and written: $ \lim_{x \to p} f(x) = L $,
or alternatively, say "$f(x)$ tends to $L$ as $x$ tends to $p$", and written: $ f(x) \to L \;\;$ as $\;\; x \to p$,
if the following property holds:
For every real $\epsilon > 0$, there exists a real $\delta > 0$ such that for all real $x \; $, $0 < x - p < \delta$ implies $f(x) - L < \epsilon$.
Or, symbolically:
$(\forall \epsilon > 0 ) \, (\exists \delta > 0) \, (\forall x \in R) \, (0 < |x - p| < \delta \implies |f(x) - L| < \epsilon)$.

