$(\delta,\varepsilon)$ 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.
$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$
 A: You need to prove that $\forall \varepsilon > 0$ $\exists \delta > 0$ : $|x-2|<\delta \implies |\dfrac{x+1}{x+2}-\dfrac{3}{4}|<\varepsilon$
You've gotten a long way now that you have that $|\dfrac{x+1}{x+2}-\dfrac{3}{4}|=\dfrac{1}{4}|\dfrac{x-2}{x+2}|$.
What you want is an expression of the form $C|x-2|$ since it is then easy to pick $\delta = \dfrac{\varepsilon}{C}$.
This is easily achievable by saying that we assume that $\delta < 1$. What is now the maximum value of $\dfrac{1}{4|x+2|}$? Well $x$ would be between 1 and 3 by rewriting $|x-2|<\delta$, so the expression would at most be equal to $\dfrac{1}{12}$. Now, you have all the necessary tools to pick $\delta$ when given an $\varepsilon$ by letting $\delta = \min\{1,
12\varepsilon\}$.
A: 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.
A: Using what you have written:
$$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$$
as $x \to 2$, from either above or below, $x-2\to0$ and therefore is certainly $\lt \epsilon$.
If we write $2+\delta$ instead of $x$ we get: $${1 \over 4}|{2+\delta-2 \over 2+\delta+2}|={1 \over 4}|{\delta \over 4+\delta}|$$
Given $\epsilon\gt0$, we need ${1 \over 4}|{\delta \over 4+\delta}|\lt\epsilon$. Assuming $\delta$ is positive (the negative case is not much different), we therefore need  ${\delta \over 4+\delta}\lt4\epsilon$.
A: 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.
