Prove that $\lim \limits_{x \to 5}\left(4x^2-7\right)=93$

Question

So I first need to determine the limit and then prove it:

$\lim \limits_{x \to 5}\left(4x^2-7\right)$

So $L=93$

And thus $\left|f(x)-L\right|=\epsilon$ and $\left|x-c\right|=\delta$

Plugging in the values...

$\left|\left(4x^2-7\right)-93\right| \lt \epsilon$, which when factored gives $4(x+5)\left|x-5\right|\lt \epsilon$

So does this mean:

$\left|x-5\right|\lt 4(x+5)\delta$

Answer 1

You should get rid of $x+5$, as it is a function of $x$. In order to do this, we fix an arbitrary delta say 1 and take minimum from 1 and a new delta which will be obtained. More precisley we already know that $|x-5|<\delta$. Now put $\delta=1$ and $9<x+5<11$ is obtained. Let us go back to our limit. One may see that $4(x+5)(x-5)<4(x-5)\times 11=44(x-5)<44\delta$ and so we have to have $\delta<\frac{1}{44}\epsilon$. Hence we can conclude that $\delta\leq min\{1,\frac{1}{44}\epsilon\}$

Answer 2

the next step.

Find $\delta$ such that when $|x-5|<\delta, |4(x + 5)(x-5)| < \epsilon$

let $\delta = \min (1, \frac {\epsilon}{44})$

Why 44?

Suppose $x-5 = \delta, |4(x + 5)(x-5)| = \delta \le 1 \implies |4(\delta + 10) \delta| = 4 \delta^2 + 40 \delta$

If $\delta \le 1$ then $|4(x + 5)(x-5)| < 44 \delta$

$\delta = \min (1, \frac {\epsilon}{44}) \implies |4(x + 5)(x-5)| < \epsilon.$