$(\delta,\varepsilon)$ 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.
$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$

Let $\varepsilon>0$ be given. Consider $|\dfrac{x+1}{x+2}-\dfrac{3}{4}|=|1-\dfrac{1}{x+2}-\dfrac{3}{4}|=|\dfrac{1}{4}-\dfrac{1}{x+2}|$
We want to find a $\delta>0$ such that whenever $|x-2|<\delta$ , the above expression $<\varepsilon$.
So start with $|\dfrac{1}{4}-\dfrac{1}{x+2}|<\varepsilon\iff|\dfrac{x-2}{4(x+2)}|<\varepsilon\iff|x-2|<4|x+2|\varepsilon$
When $|x-2|<\delta$, we can say that $2-\delta<x<2+\delta\implies|x+2|<4+\delta<M$ where $M$ is a uniform upper bound. Thus, we have, $|x-2|<4|x+2|\varepsilon<4M\varepsilon$ i.e. $|x-2|<4M\varepsilon$. Choose $\delta=4M\varepsilon$. Then, we have got our $\delta$.
Now check that this $\delta$ works.
