Question:
Prove that if $\displaystyle \lim_{x \to a} f(x) = L$ then there is a number $\delta > 0$ and a number $M$ such that $|f(x)|<M$ if $0 < |x - a|< \delta$.
This means:
For every $\epsilon > 0$ and some $\delta > 0$
$|f(x) - L| < \epsilon$ for $|x - a| < \delta$
By the Reverse Triangle Inequality:
$\bigg| |f(x)| - |L| \bigg| < |f(x) - L| < \epsilon$ for $|x - a| < \delta$
$\bigg| |f(x)| - |L| \bigg| <\epsilon$ for $|x - a| < \delta$
$-\epsilon < |f(x)| - |L|< \epsilon$ for $|x - a| < \delta$
$\therefore |f(x)| < \epsilon + |L|$ for $|x- a| < \delta$
Hence we let $M = \epsilon + |L|$
Is this how it is supposed to work? SO that $M \ne L$ but $|f(x)| < M$?
Thanks