Proving the function g is continuous without using continuous function theorem

Question

Suppose that $f$ is continuous at $a$, $f(a) = 0$, and $g(x) = 5f(x)^{2}$. Give an $\epsilon - \delta$ proof that g is continuous at a. *Do not use the continuous function theorem.

I am confused as to how I can choose a $\delta$.

Answer 1

$g(a)=0$, hence you have to prove that $\forall\epsilon>0\;\;\exists\delta>0$ such that $$ |g(x)|=5f(x)^2<\epsilon\;\;\;\forall|x-a|<\delta. $$ Since $f$ is continuous at $a$ by hypotesis, for every $\epsilon>0$ there exists $\delta$ such that $$ |f(x)|<\sqrt{\epsilon/5}\;\;\;\forall|x-a|<\delta $$ which allows you to conclude.