Given $|g(x)|\leq M|x-2|$, prove that g is continuous at $x=2$

Question

Given $|g(x)|\leq M|x-2|$, prove that g is continuous at $x=2$

I am not sure whether the way I solve it is correct or not.

Given any $\epsilon>0$, take $\delta=\frac{\epsilon-|g(2)|}{M}$

$$|x-2|<\delta \implies|g(x)-g(2)|\leq |g(x)|+|g(2)|\leq M|x-2|+|g(2)|<\epsilon$$

Answer 1

Mostly right, though the way $g(2)$ just disappears near the end looks rather suspect. It would be safer to start by proving explicitly that $g(2)=0$ so you can discard it early on.