# Verify from this proof of $\lim x^2$

Preliminary Analysis:

$$|x^2-4| = |(x-2)(x+2)| = |x-2|<\frac{\varepsilon}{|x+2|}$$

Since $$\delta$$ can not depend on $$x$$ then

choose a small $$\delta_i$$ such that $$\delta_i<\frac{\varepsilon}{|x+2|}$$ where $$\delta_i$$ does not depend on $$x$$

Impose $$|x-2|≤ 1$$ then

$$-1 ≤x-2 ≥1$$

$$= 1 ≤x ≥3$$

$$= 3 ≤x+2 ≥5$$

$$= -5<3 ≤x+2 ≥5$$

$$= |x+2| ≤5$$

put $$\delta_i = \frac{\varepsilon}{5}$$

then $$\delta_i = \frac{\varepsilon}{5}<\frac{\varepsilon}{|x+2|}$$

Now we have $$|x^2-4| = |(x-2)(x+2)| < 5.|x-2| = |x-2|<\frac{\varepsilon}{5}$$

Formal proof:

$$\lim_{x\to 2}f(x)=x^2\iff (\forall\varepsilon>0)(\exists\delta>0):0<\lvert x-2\rvert<\delta = \min({1,\frac{\varepsilon}{5}})\implies\bigl\lvert x^2-4\rvert = |(x+2)(x-2)| < 5.|x-2|<5 \frac{\varepsilon}{5} = \varepsilon$$