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.
 A: To solve these types of problems, you usually start with what you want to show and work backwards to try to figure out a $\delta$ that will work.
So let $\epsilon > 0$ be arbitrary.  At the end of the day, we want $\left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon$.  Let's try to manipulate the left hand side in ways we are allowed to do and hope we get something useable (you have to guess and try things and see if they lead you anywhere).
Ok, well one thing we can do is try to add the fractions up, and factor or cancel or do any other algebra that seems obvious to try.  So we have 
$$\begin{split} \left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| 
&= \left | \dfrac{(x + 2)(a^{2} - 5) - (a + 2)(x^{2} - 5)}{(a^{2} - 5)(x^{2} - 5)} \right | \\ 
&=\left | \dfrac{xa^{2} - 5x + 2a^{2} - 10 - ax^{2} + 5a - 2x^{2} + 10}{(a^{2} - 5)(x^{2} - 5)} \right | \\ 
&=\left | \dfrac{xa^{2} -ax^{2} - 5x + 5a + 2a^{2} - 2x^{2} }{(a^{2} - 5)(x^{2} - 5)} \right | \\ 
&= |x-a| \left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right |.  \end{split}$$
Ok, well after trying some algebra, we ended up with a factor of $|x - a|$ which is good!  Usually, if you end up with something like $M|x-a|$ for some number $M$, then you can let $\delta = \epsilon/M$, because then we would get $M |x -a| \leq M (\epsilon/M) = \epsilon$ (if $|x - a| < \delta$).
The only problem is the other factor multiplied by our $|x - a|$ isn't just a number -- it depends on $x$.  We don't want any $x$ presence in the other factor.  So let's do a standard trick which you already seem to know.  Let's assume without loss of generality that $\delta < $ some number (do you understand why we can do this?).  Once you understand why we can do this, let's talk about how we do this: if there are no denominators to worry about, we can usually pick $\delta < 1$.  But in this case, we have $x^{2} - 5$ in the denominator.  We don't want this to be $0$ since otherwise the expression would be undefined!  So we want to pick a number $\delta$ so that $x$ can't possibly be $\pm \sqrt{5}$.  Since $a$ is either in $(-\sqrt{5}, \sqrt{5})$ or otherwise is in $(-\infty, -\sqrt{5}) \cup (\sqrt{5},\infty)$ (because $a$ cannot equal $\pm \sqrt{5}$), there is some number $\rho > 0$ so that $(a - \rho, a + \rho)$ doesn't contain $\pm \sqrt{5}$.  Pick $\delta$ small enough so that if $|x - a| < \delta$, then $x \in (a - \rho, a + \rho)$.  So basically, pick $\delta \leq \rho$, where $\rho$ is the fixed number described above.  This will ensure $x$ cannot be $\pm \sqrt{5}$.
So then if $\delta \leq \rho$, we have $|x - a| < \delta$ implies $|x - a| < \rho$ so that $-\rho + a < x < \rho + a$.  Let $M = \max\{|-\rho + a|, |\rho + a|  \}$.  So $-M < x < M$, which means $|x| < M$.
Now, try to convince yourself of why the following is true (based on $|x|<M$):
$\left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right | \leq \dfrac{|xa| + 5 + 2|x| + 2|a| }{|(a^{2} - 5)(x^{2} - 5)|} \leq \dfrac{M|a| + 5 + 2M + 2|a| }{|(a^{2} - 5)(0^{2} - 5)|} = \dfrac{(M + 2)|a| + 5 + 2M}{5|(a^{2} - 5)|}$
Let's call the last expression on the right hand side the letter $K$. It's just a constant number that doesn't depend on $x$.
Then, based on all of the work above, if we let $\delta = \min\{\rho, \epsilon/K\}$ (we are taking the minimum with $\rho$ because we assumed  $\delta \leq \rho$ above), then we get $|x -a| < \delta \implies$ $$\left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right|  = |x-a|\left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right | < |x-a| K < (\epsilon/K)K = \epsilon$$
A: $$
\begin{align}
\left |\frac{x+2}{x^2-5}-\frac{a+2}{a^2-5} \right| & = \left | \frac{(x + 2)(a^2 - 5) - (a + 2)(x^2 - 5)}{(a^2 - 5)(x^2 - 5)} \right | \\[10pt]
& = \left | \frac{ax+2a+2x-5}{(a^2 - 5)(x^2 - 5)} \right | \cdot \underbrace{ |x-a|}_\text{important}.
\end{align}
$$
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 $0$.
