<s>[INST] Using the $\epsilon$-$\delta$ definition of the limit, evaluate $ \lim_{x \to a} \frac{x+2}{x^2-5}$ 
Using the $\epsilon$-$\delta$ definition of the limit, evaluate $$ \lim_{x \to a} f(x)$$ where $$f(x) = \frac{x+2}{x^2-5}$$

Attempt
We need to show that $\forall \epsilon, \exists \delta$ such that $$0 < |x-a| < \delta \quad \implies \quad \left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon.$$ I don't see an easy way to do this algebraically so I am going to say let $|x-a| < 1$ and then $a - 1 < x < a+1$. We then have $\left | \dfrac{a+1}{(a-1)^2}-\dfrac{a+2}{a^2-5}\right | < \epsilon$. Then I am not sure how to simplify it to prove the statement.
 [/INST] $$
\begin{align}
\left |\frac{x+2}{x^2-5}-\frac{a+2}{a^2-5} \right| & = \left | \frac{(x + 2)(a^2 - 5) - (a + 2)(x^2 - 5)}{(a^2 - 5)(x^2 - 5)} \right | \\[10pt]
& = \left | \frac{ax+2a+2x-5}{(a^2 - 5)(x^2 - 5)} \right | \cdot \underbrace{ |x-a|}_\text{important}.
\end{align}
$$
Pulling out $|x-a|$ is predictably possible because the whole expression will of course be $0$ when $x=a$.
And $|x-a|$ is the thing that you will make $<\delta$.
Now you need to get a bound on the first factor that does not depend on $x$, so that you have
$$
\left |\frac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| \le (\text{constant}\cdot|x-a|),
$$
where "constant" means "not depending on $x$".  You use the prior decision to make $\delta\le 1$ to assure that $x$ is between $a\pm\delta$, and thus to make the "constant" independent of $x$ [but see the PS below].
PS: As "Ian" points out in comments below, rather than just $\delta\le 1$ we need $\delta$ small enough to keep $a\pm\delta$ away from the two square roots of $5$, i.e. $\pm\sqrt 5$, in order to keep the denominator away from $0$.
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