Dealing with 2 absolute values in one inequality.

Question

Could anyone tell me how $|f'(x)- L| < |L|/2$ give us $|f'(x)| \gt |L|/2$ ? and $L \ne 0$.

Answer 1

A simple proof is to use

$$|a - b| \ge |a| - |b|$$

$$|L|/2 \gt |f'(x) - L| = |L - f'(x)| \ge |L| - |f'(x)|$$

Thus

$$|f'(x)| + |L|/2 \gt |L| \implies |f'(x) \gt |L| - |L|/2 = |L|/2$$