<s>[INST] $(\delta,\varepsilon)$ Proof of Limit I wish to prove that $\lim_{x\to 2} {x+1 \over x+2} = {3 \over 4} $.
The $(\delta,\varepsilon)$ limit definition in this case is:
$\forall \epsilon >0, \exists \delta >0$ such that $0<|x-2|<\delta \Rightarrow |{x+1 \over x+2} - {3 \over 4}| < \epsilon.$
Thus, I need to provide a $\delta$, which is a function of $\epsilon$ in order to satisfy the above definition.
I am having a bit of difficulty finding an inequality to continue my work below.
$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$
 [/INST] You want to prove
$$\lim_{x\to 2}\frac{x+1}{x+2}=\frac{3}{4}.$$
By the definition of limit you need to find, for all $\epsilon>0$, a $\delta$ such that $0<|x-2|<\delta$ implies $$\left|\frac{x+1}{x+2}-\frac{3}{4}\right|<\epsilon.$$ Your best bet is to work back from the complicated expression. We can simplify,
$$\left|\frac{4(x+1)-3(x+2)}{4(x+2)}\right|=\left|\frac{x-2}{4(x+2)}\right|=\frac{1}{4}\left|\frac{x-2}{x+2}\right|=\frac{1}{4}\frac{|x-2|}{|x+2|}<\epsilon.$$ Look what just appeared in the numerator - it's the $|x-2|$ you needed to get a handle on.
Rearranging we have $$|x-2|<4|x+2|\epsilon.$$
Suppose $|x-2|<1$ (which we suppose because we want to be "close" to 2, here within a distance of 1), then $$-1<x-2<1,$$ or adding 4 to both sides, $$3<x+2<5,$$ from which we can write $|x+2|<5$.
Substituting, we get
$$|x-2|<(4\cdot 5)\epsilon,$$ or $$|x-2|<20\epsilon.$$ So let $\delta = \min\{1,20\epsilon\}$.

Now let's check this works: For all $\epsilon>0$, let $|x-2|<\delta=\min\{1,20\epsilon\}$. We can see from this inequality that $\delta\leq 20\epsilon$ holds true regardless of what the minimum turns out to be. Hence, we have
$$|x-2|<20\epsilon\iff \frac{1}{4}|x-2|<5\epsilon.$$
Rewriting this as $$\frac{1}{4}\frac{|x-2|}{|x+2|}|x+2|<5\epsilon,$$ and then using the fact that
$$|x-2|<1\iff -1<x-2<1\iff 4<x+2<5\implies|x+2|<5,$$
then we deduce that $$\frac{1}{4}\left|\frac{x-2}{x+2}\right|<\epsilon,$$ as required.
</s>