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$