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?
Best Answer
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$$|x-2|<\delta\implies\bigl|(x-2)(12x+21)\bigr|<\frac\varepsilon{57}\times57=\varepsilon.$$