How to calculate limits of functions rigorously
Andrew Henderson 
I am using the following definition of the limit of a function "
Definition: Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. Suppose that $E \subseteq X$. Let $f : E \to Y$ and suppose that $p$ is a limit point of $E$. We say that $$\lim_{x \to p} f(x) = q$$ if there exists a point $q \in Y$ with the property that for all $\epsilon > 0$ there exists a $\delta > 0$ such that $$0 < d_X(x, p) < \delta \implies d_Y((f(x), q) < \epsilon$$ for any $x \in B_{(X, d_x)}(p, \delta)$
Suppose now I had the following function $f(x) : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ defined by $$f(x) = \frac{x^2}{x^4 + x^2}$$ and I wanted to prove $$\lim_{x \to 0} \ f(x) = 1.$$ Question 1: What is the easiest way to prove this?
In attempting to prove this on my own I simplified $f(x)$ to get that $$f(x) = \frac{x^2}{x^4 + x^2} = \frac{1}{x^2+1}$$, thus $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x^2 + 1}$$
Now I could try and prove it directly using the $\epsilon-\delta$ definition above, but I never see anyone actually calculating limits like this in practice (I'm assuming there's a theorem that allows us to prove the above in a much simpler way).
Indeed there is a theorem that looks like it could be useful here
Theorem: Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, let $E \subseteq X$ and let $f : E \to Y$ . Assume that $p$ is a limit point of $E$, then $f$ is continuous at $p$ if and only if $$\lim_{x \to p} f(x) = f(p)$$
Now note that the function $g : \mathbb{R} \to \mathbb{R}$ defined by $$g(x) = \frac{1}{x^2+1}$$ is continuous on $\mathbb{R}$ and that $0$ is a limit point of $\mathbb{R}$ hence $$\lim_{x \to 0} g(x) = g(0) = 1$$
and this looks like a very promising result that we could use to prove that $\lim_{x \to 0} f(x) = 1$, however there's one big problem, $f$ is defined on $\mathbb{R} \setminus \{0\}$ and since $0$ is not in the domain of $f$, it follows that $f$ can't be continuous at $0$. So we can't make use of the above theorem directly to prove that limit for $f$.
Question 2: Is there any way I can use the above theorem to prove that $\lim_{x \to 0} \ f(x) = 1$?
$\endgroup$2 Answers
$\begingroup$Note that $\lim_{x\to 0} f(x) = \lim_{x\to 0} g(x) $ because in the definition of the limit we do not require the function to be defined at the limit point: $$0 < d_X(x, p) < \delta \implies d_Y((f(x), q) < \epsilon$$
We see that from $0 < d_X(x, p)$ (we look at points $x$ that are close to $p$ but are not $p$).
Let us show this rigorously. Assumme $\lim_{x\to 0} g(x)$ exists. That is, there exists a point $q\in\mathbb{R}$ such that for every $\epsilon>0$ there exists a $\delta>0$ such that for all $x\in \mathbb{R}$, $0<|x|<\delta \implies |g(x)-q|<\epsilon$. When $0<|x|<\delta$ we have that $g(x) = f(x)$. Thus, we have $0<|x|<\delta \implies |f(x)-q|<\epsilon$. From the definition of the limit we can say $\lim_{x\to 0} f(x)=q$ which in fact implies $\lim_{x\to 0} f(x) = \lim_{x\to 0} g(x) $.
Thus, the method you have presented is a valid method to calculate the limit (and the easiest one - for continues functions, the easiest way to find the limit is by this definition).
$\endgroup$ 2 $\begingroup$Regarding your Question 1, you can do a substitution $x=1/u$ to find $ \lim_{u \rightarrow \infty } \frac{u^2}{1+u^2} $ instead.
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