Proof that $\operatorname{lim}_{x \rightarrow x_0} x^2 = x_0^2$

Question

Let $f:\mathbb R \rightarrow \mathbb R$ be a function defined by $f(x) = x^2$ for all $x \in \mathbb R$. I want to show using an epsilon-delta proof that $lim_{x \rightarrow x_0} x^2 = x_0^2$ for all $x_0 \in \mathbb R$. I let $\epsilon \gt 0$, and a textbook I'm reading says to choose $\delta = min \{1, \frac{\epsilon}{2|x_0| + 1}\}$. Why?

I eventually end up getting that $|f(x) - x_0^2| < \delta(2|x_0| + 1)$ for all $x \in \mathbb R$ with $0 < |x - x_0| < \delta$. I'm a bit confused on why delta was chosen that way. Can someone please explain?

Thanks in advance.

Answer 1

Without loss of generality you may assume $\delta < 1$

If $|x-x_0|<\delta$ then you have $|x-x_0|<1$ that is $$|x+x_0|=|x-x_0+2x_0|\le |x-x_0|+|2x_0|<1+2|x_0|$$

Now you can proceed with the rest of the proof.