A different way to prove that $\lim\limits_{x \to 2} x^{2} = 4$?

Question

I see many people choosing $\delta = \min (1, \frac{\epsilon}{5}).$ But can't we just choose $\delta = \frac{\epsilon}{|x+2|+1}$? Then we have, under the assumption that $0<|x-2|<\delta,$ that $$|x^{2}-4|=|(x-2)(x+2)|=|x-2||x+2| < \frac{\epsilon}{|x+2|+1}|x+2| < \epsilon.$$

Answer 1

Just like it was mentioned in the comments, your $\delta$ can't depend on $x$. I'm not sure if from the title you want a different way to actually prove the limit or if you just want to choose $\delta = \frac{\epsilon}{|x+2|+1}$, so feel free to say my answer doesn't help or downvote it. But a different way is letting $\delta = \sqrt{\epsilon+4} - 2$. Using your $0 < |x-2| < \delta$, we get that $$ \eqalign{ -\delta &< x-2 < \delta \cr -\delta +4 &< x+2<\delta+4 \cr -\delta - 4 &< x + 2 < \delta + 4, } $$ which implies that $|x+2| < \delta + 4$. Then

$$|x^2-4| = |x+2|\cdot |x-2| < \delta(\delta+4) = (\sqrt{\epsilon+4}-2)(\sqrt{\epsilon+4}-2+4) = \epsilon.$$