How do I choose $\delta$ in the $\epsilon-\delta$ proofs of limits. While I understand how $\delta$ is chosen when a funtion is nice and easy, linear function. But I'm having trouble understanding how the $\delta$ is selected for quadratic function or rational functions.
For instance, 
I have to prove that, $\lim\limits_{x \to 2} \dfrac{x^3-4}{x^2+1} = \dfrac{4}{5}$
So let $f(x)$ be the given function,
Fixing an $\epsilon>0$, and simplifying $|f(x)- 4/5|$ I get,
$\left| f(x)- \dfrac{4}{5} \right| = \dfrac{|5x^3 +6x+12|}{5(x^2+1)} \cdot |x-2| $
Now I understand I need to get rid of everything except $|x-2|$ possibly by replacing the entire thing by a number.
But I have no clue as to how you choose this number. What should be my thought process now? 
Thanks! 
 A: A convenient way would be to require $\delta<1$, so that we can take 
$$
1<x<3
$$
Then, the smallest $x^2+1$ can be is $2$ and the largest the increasing $5x^3+6x+12$ can be is $5(3)^3+6\cdot 3+12=165$, so 
$$
\frac{5x^3+6x+12}{5(x^2+1)}\leq \frac{165}{10}=16.5
$$
now you can take 
$$
\delta=\frac12\min \{ 1,\frac{\epsilon}{16.5} \}
$$
Note, there is no canonical choice for $\delta>0$, meaning lots of choices will work. 
A: Your thought process might be something like
I know I can make the $|x-2|$ small just by a direct choice of $\delta$. But that other factor might make the right side big. Can I bound the first factor on the right, so that I can be sure it won't do too much damage? Well, it's a fraction. Fractions are large when their denominators are small. The denominator of the factor is always at least $5$, so the fraction is no larger than $1/5$ of the numerator. Now the numerator of the fraction is an increasing function near $x=2$, so as long as $x < 10$ the numerator will be less than $6000$. (I could get a much better bound than that with a smaller $x$ and more arithmetic, but why bother?) 
Now I know that when $x < 10$ the right side is  bounded by $(6000/5)|x-2|$. That tells me that as long as $\delta < 10$ and $\delta < (5/6000)\epsilon$ I'll have the bound I need on the value of the function.
