Proving $\lim_{x \to 2}x^2 = 4$ true

Question

I have this limit: $\lim_{x \to 2}x^2 = 4$ and I want to prove it, so I solve $|f(x)-l|<\epsilon$

$|x^2-2|<\epsilon$

$2-\epsilon<x^2<2+\epsilon$

and here I don't know what to do, can you help me?

Answer 1

Given $\epsilon>0$, you look for $\delta $ such that

$$|x-2|<\delta \implies |\color {green}{x+2} ||x-2|<\epsilon $$

As $x $ goes to $2$, it means that $x $ is not far from $2$, so we can assume that $$|x-2|<\color {red}{1} $$

$$\implies -1 <x-2 <1 $$ $$\implies 3 <\color {green}{x+2} <5$$

$$\implies$$ $$ |x^2-4|=|\color {green}{x+2}||x-2|<5|x-2|.$$

thus

$\Bigl (5|x-2|<\epsilon $ and $|x-2|<\color {red}{1} $

$\implies |x^2-4|<\epsilon\Bigr) $.

To satisfy this last condition, we need $|x-2|<\frac {\epsilon}{5} $ and $|x-2|<\color {red}{1} $

hence we will take $\delta=\min (\frac {\epsilon}{5},\color {red}{1}) $.