If $\lim_{z\rightarrow z_0} g(z)=B\neq 0$ then $|g(z)|>\frac{1}{2}|B|$

Question

My teacher said that if $\lim_{z\rightarrow z_0} g(z)=B\neq 0$ then $|g(z)|>\frac{1}{2}|B|$. I am trying to prove this, below is my attempt

For all $ \epsilon$, $\exists \delta>0$ such that

$\epsilon>|g(z)-B|\geq||g(z)-|B||>||g(z)-\frac{1}{2}|B||$

I am not sure what to do next.

Answer 1

Let $\epsilon = \dfrac{||B||}{2} \implies \exists \delta > 0$, such that if $0 < ||z-z_0||< \delta\implies ||B||-||g(z)||\le ||g(z) - B|| < \dfrac{||B||}{2} \implies ||B|| - ||g(z)|| < \dfrac{||B||}{2}\implies ||g(z)|| > \dfrac{||B||}{2}$.