Let $f:\mathbb{R}\to\mathbb{R}$ be such that $\lim_{x\to p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $\varepsilon-\delta$ formalism, that $$\lim_{x\to p} [f(x)]^2 = L^2$$
My work:
$$\lim_{x\rightarrow p} [f(x)]^2 = \lim_{x\to p}f(x) \cdot\lim_{x\to p} f(x) = L \cdot L = L^2$$
But that seems too easy, and doesn't involve $\varepsilon$ or $\delta$. Any other approaches? Please and thank you.
Perhaps a more appropriate phrasing for this question would be:
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be such that $\lim_{x\rightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $\epsilon-\delta$ definition of a limit, that $$ \lim_{x\rightarrow p}[f(x)]^2 = L^2 $$
We aren't just looking for some $\epsilon$'s and $\delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$\lim_{x\rightarrow p}f(x)=L$ if and only if $\forall\epsilon>0$ $\exists\delta>0$ such that if $0<|x-p|<\delta$ then $|f(x)-L|<\epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$\lim_{x\rightarrow p}[f(x)]^2=L^2$ if and only if $\forall\epsilon>0$ $\exists\delta>0$ such that if $0<|x-p|<\delta$ then $|f(x)^2-L^2|<\epsilon$.
Please note that the $\epsilon$'s and $\delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.