Epsilon-Delta proof of a quadratic limit I just started learning about limits and their definition. I'm practicing with some proofs that involve the epsilon-delta definition of a limit.While I absolutely have no problem with linear functions, I really don't know how to prove quadratic limits. This is the original limit that I'm trying to prove:
$\lim \limits_{x\to2} (12x^2-3x+8)=50$
I started with the epsilon- delta definition.
Given an $\epsilon$  > 0, there is always a $\delta$ >0 such that:
If $|x-2|<\delta$,then $|12x^2-3x+8-50| = |12x^2-3x-42|<\epsilon$
I found out that $12x^2-3x-42 =(x-2)(12x+21)$. So, by starting from the hypothesis:
$|x-2|<\delta$
$-\delta<x-2<\delta$
$-\delta(12x+21)<(x-2)(12x+21)<\delta(12x+21)$
$-\delta(12x+21)<(12x^2-3x-42)<\delta(12x+21)$
$|12x^2-3x-42|<\delta(12x+21)$, this form is very similar to $|12x^2-3x-42|<\epsilon$, so this suggests me
that i should take $\delta=\frac{\epsilon}{12x+21}$, so that $\delta(12x+21)=\epsilon$
Now I have a big problem. $\delta$ cannot depend on x, so I must find a bound, but I do not know how.
I thought that maybe I could first consider values of x that are greater than 2
if $x>2$
$12x>24$, which means that $12x+21>45$
If $12x+21>45$,then $\frac{\epsilon}{12x+21}<\frac{\epsilon}{45}$, so $|x-2|<\delta=\frac{\epsilon}{12x+21}<\frac{\epsilon}{45}$, which implies that $|x-2|<\frac{\epsilon}{45}$. So a new choice for $\delta$ could be $\frac{\epsilon}{45}$, but it is valid only for $x>2$.
At this point I thought that I should also consider values of x that are less than 2.
So for $x<2$, I have that $12x+21<45$, so now $\frac{\epsilon}{12x+21}>\frac{\epsilon}{45}$,,so $|x-2|<\frac{\epsilon}{12x+21}>\frac{\epsilon}{45}$. I'm blocked right here because I cannot find a new delta for x < 2. How should I proceed?
 A: Take $\varepsilon>0$. Note that\begin{align}|x-2|<1&\iff1<x<3\\&\iff12<12x<36\\&\iff33<12x+21<57\\&\implies|12x+21|<57.\end{align}So, take $\delta=\min\left\{1,\frac\varepsilon{57}\right\}$. Then$$|x-2|<\delta\implies\bigl|(x-2)(12x+21)\bigr|<\frac\varepsilon{57}\times57=\varepsilon.$$
A: Your goal is to prove that the following is true:

Given an $ \epsilon> 0$, there is always a $\delta>0$ such that:
$$ \text{If } |x−2|<\delta \text{ ,then } |12x^2-3x+8-50| = |12x^2-3x-42|<\epsilon$$

First solve in $x$ the following parametric inequality: $|12x^2-3x-42|<\epsilon$.
This leads to a solution set $S_{\epsilon}$ which depends on the value of $\epsilon$.
To finish the job you have to prove that, for every positive value of the parameter $\epsilon$, inside the corresponding set$S_{\epsilon}$ lies every point satisfying $|x−2|<\delta$ provided $\delta$, which depends (in general) on $\epsilon$, is sufficiently small.
A: You want to solve these problems backwards, starting with your destination and then reverse engineering to figure out what limitations you need on $\delta$ in terms of $\varepsilon$.
You want $\vert (12 x^2-3x+8)-50 \vert = \vert 12x^2-3x-42 \vert = 3 \vert 4x^2-x -14 \vert \lt \varepsilon$, so you want $\vert 4x^2 - x -14 \vert \lt \frac \varepsilon 3$ near $x=2$.
That means you want $\vert(4x+7)(x-2)\vert = \vert 4x+7 \vert \vert x-2 \vert  \lt \frac \varepsilon 3$ near $x=2$.
That means you want: $$\vert x-2 \vert \lt \frac {\varepsilon}{3\vert 4x+7 \vert}$$
near $x=2$.  If $\vert x-2 \vert \lt 1$, then $x \lt 3$ and $\vert 4x+7 \vert \lt 19$.
That means if you choose $\delta = \min(1, \frac {\varepsilon}{57})$ and work backwards through what I've written above, you'll see that $\vert x-2 \vert \lt \delta \Rightarrow \vert (12x^2-3x+8)-50 \vert \lt \varepsilon$, as required.  You'll get it down to a product of two terms, one of which is guaranteed to be less than $\frac {\varepsilon} {57}$ because $\delta \lt \frac{\varepsilon}{57}$ and the other of which is guaranteed to be less than $57$ because $\delta \lt 1$.
A: I find it easier to directly start trying to make an inequality chain, assuming $|x-2|<\delta$:
$$
|(12x^2-3x+8)-50| = |12x+21||x-2| < |12x+21|\delta
$$
Now we want to put some constant bounds on $|12x+21| = |12(x-2)+45|$ whenever $|x-2|<\delta.$ We can do this by requiring $\delta<1$ so that $|12(x-2)+45| < 57$. We can then take $\delta \leq\frac{\epsilon}{57}$.
Thus take $\delta = \min(1, \frac{\epsilon}{57})$ and we will have
$$
|(12x^2-3x+8)-50| = |12(x-2)+21||x-2| < 57\delta \leq 57\cdot\frac{\epsilon}{57} = \epsilon.
$$
