Proving $0<|x-a|<\delta$ =>$|g(x) > |M|/2$ (Epsilon-Delta problem)

Question

As the title says I have to prove that:

if $\lim_{x\to a}$ $g(x) = M$

Where $M \not= 0$,

Show that there exist a number $\delta >0$ such that:

$0<|x-a|<\delta$ =>$|g(x)| > |M|/2$

How do I even approach this problem?

Thank you in advance

Answer 1

By the definition of the limit, we have that for all $\epsilon>0$ there exists $\delta>0$ such that when $|x-a|<\delta$ we have $|g(x)-M|<\epsilon.$ Or $M-\epsilon<g(x)<M+\epsilon.$ Assuming $M>0$ take $\epsilon =M$ then there exists $\delta>0$ such that when $|x-a|<\delta$ then we have that $g(x)>M/2$.

Answer 2

The intuition behind the statement $$\lim_{x \to a}g(x)=M$$ is the following. When your point $x$ is really close to $a$, $g(x)$ is really close to $M$. In particular, if $M \neq 0$ you know there exist an $\epsilon_0 >0,$ maybe very small, such that $$\vert M \vert -\epsilon_0 > \dfrac{\vert M \vert}{2}$$ which implies that $\epsilon_0 < \vert M \vert /2.$ Moreover, by the definition of limit, you know that for all $\epsilon >0$, there exist $\delta >0$ such that if $0 < \vert x-a \vert <\delta,$ then $ \vert g(x) - M \vert =\vert g(x) - \lim_{x \to a}g(x) \vert < \epsilon.$

Therefore, you can pick $\delta_0 > 0$ such that if $0 < \vert x-a \vert <\delta_0$, then $$\vert M \vert - \vert g(x) \vert < \vert g(x) - M \vert <\epsilon_0 < \vert M \vert /2.$$ This implies that $\vert g(x) \vert > \vert M \vert/2.$