How is the inequality reversed in this limit proof?

Question

In the above proof, how did they go from $|g(x)-B|<|B|/2$ to |g(x)|>|B|/2? Wouldn't that only work if $B$ is positive?

Answer 1

Intuitively, if $g(x)$ is closer to $B$ than $0$, then thinking about distances from $0$, $g(x)$ must be farther than $B/2$ from $0$. This then implies that $|g(x)| > |B|/2$.

You can prove this formally using the reverse triangle inequality, i.e. ($||a| - |b|| \leq |a - b|$). Observe that

$$|g(x)| = |B - (g(x) - B)| \geq ||B| - |g(x) - B|| \geq |B| - |g(x) - B| > \frac{|B|}{2}.$$