Suppose that $\lim\limits_{x \rightarrow p}f(x) = L$. Prove that there are $\delta>0$ and $M>0$ such that,
$0<|x-p|<\delta \implies |f(x)| \le M$
That's how I did:
Given that $\lim\limits_{x \rightarrow p}f(x) = L$, then for any $\epsilon>0$ there exists $\delta>0$ such that
$0<|x-p|<\delta \implies L-\epsilon <f(x)< L+\epsilon$
Now let $M=|L|+\epsilon$. We know that $L \le |L|$ and $-|L| \le L$, then
$f(x)<L+\epsilon \le |L|+\epsilon=M$ and $-M=-|L|-\epsilon \le L-\epsilon<f(x)$
Then $-M<f(x)<M$.
This way, there is $\delta>0$ and $M>0$ such that
$0<|x-p|<\delta \implies |f(x)| \le M$
Is this correct ?
