# Real Analysis Proof

• Nov 16th 2008, 10:35 AM
Nightfly
Real Analysis Proof
Please can someone help me, I'm hopeless with proofs and this is driving me crazy.

If $\displaystyle lim_{x \to a} g(x) = m$ with $\displaystyle m \neq 0$ then
$\displaystyle lim_{x \to a} \frac{1}{g(x)} = \frac{1}{m}$
• Nov 28th 2008, 07:29 PM
TheEmptySet
Quote:

Originally Posted by Nightfly
Please can someone help me, I'm hopeless with proofs and this is driving me crazy.

If $\displaystyle \lim_{x \to a} g(x) = m$ with $\displaystyle m \neq 0$ then
$\displaystyle \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{m}$

There are two cases here is the first. Suppose m > 0

Since the limit exits $\displaystyle \delta_1 > 0$ such that when $\displaystyle |x-a|< \delta_1 \implies |g(x)-m|< \frac{m}{2} \implies$

$\displaystyle -\frac{m}{2}< g(x)-m < \frac{m}{2} \implies \frac{m}{2} < g(x)$

you can show something similar when m <0 together this implies that

Now choose a $\displaystyle \delta_2$ such that when

$\displaystyle |x-a|< \delta_2 \implies |g(x)-m|< \frac{m^2 \epsilon}{2}$

let $\displaystyle \delta =min\{\delta_1,\delta_2 \}$

if $\displaystyle |x-a|< \delta$

$\displaystyle |\frac{1}{g(x)}-\frac{1}{m}|=\frac{|m-g(x)|}{|m\cdot g(x)|}<\frac{\delta}{m\frac{m}{2}}< \epsilon$