How to prove that a function $f$ has a limit $l$ when $x \to x_0$, using the ε-δ definition? I'm trying to prove that $$\lim_{x\to 2}\Bigl(\frac{x+1}{x+2}\Bigr)=\frac34$$
by using the $ε-δ$ definition of the limit (without a given $ε$). 
What bugs me is that there is no apparent general way to define $δ$. I've read many solved examples, but pretty much every single one of them just seemed to pull the value of $δ$ out of thin air, without any explanation for their reasoning.  Could someone explain to me how is that $δ$ calculated? (preferably on this example, because for simple non-quotient functions I can simplify the $|f(x) - x_0| < ε$ and carry on)
Thanks in advance.
 A: Not sure if this helps, but we want $$\left |\frac{x+1}{x+2}-\frac{3}{4}\right|=\left |\frac{x-2}{4(x+2)}\right|<\epsilon$$ so lets see if we can bound $\frac{1}{4(x+2)}$. We know that $$\left |x-2\right|<\delta$$which is the same as saying $-\delta<x-2<\delta$. Adding $4$ to both sides we get $4-\delta<x+2<\delta+4$, so $$\frac{1}{4(x+2)}<\frac{1}{16-4\delta}$$ We can therefore say that $$\left |\frac{x-2}{4(x+2)}\right|<\frac{\delta}{16-4\delta}=\frac{1}{\frac{16}{\delta}-4}$$ We can set this bound equal to $\epsilon$ and we'll get $$\delta=\frac{16}{1+\frac{1}{\epsilon}}=\frac{16\epsilon}{\epsilon+1}$$
Notice that we haven't used that kind of argument you often see, like "let $|x-2|<\delta\leq 1$"
A: To find your $\delta$ work backwards from what you want
$$
\left|\frac{x+1}{x+2}-3/4 \right|<\epsilon\\
\iff \left|\frac{4x+4-3x-6}{4(x+2)} \right|<\epsilon\\
\iff \left|\frac{x-2}{x+2} \right|<4\epsilon\\
$$
now we can control $|x-2|$ using $\delta$. 
But the $|x+2|$ term we can also control, intuitively this is because we care about when $x\approx 2$, so it should be around $4$. To encode this, we see that as long as 
$$
\delta<1
$$
we get $3<x+2<5$ and helpfully 
$$
\frac{|x-2|}{|x+2|}<\delta/3
$$
making a convenient choice of $\delta$ to be 
$$
\delta=\min\left\{ 1,12\epsilon\right\}
$$
try and work through the above backwards and make sure my choice of $\delta$ works.
