Question

Evaluate the given limit:
$\underset{x\to -2}{lim}\frac{(\frac{1}{x}+\frac{1}{2})}{x+2}$
.

Answer 1

Given : $\underset{x\to -2}{lim}\frac{(\frac{1}{x}+\frac{1}{2})}{x+2}$
At $x=-2$, the value of the given function takes the form $\frac{0}{0}$
So, simplifying
$\underset{x\to -2}{lim}\frac{(\frac{1}{x}+\frac{1}{2})}{x+2}=\underset{x\to -2}{lim}\frac{\frac{2+x}{2x}}{x+2}(\because x\ne -2)$
$=\underset{x\to -2}{lim}\frac{1}{2x}$
$=\frac{1}{2(-2)}$
$=-\frac{1}{4}$
$\therefore \underset{x\to -2}{lim}\frac{(\frac{1}{x}+\frac{1}{2})}{x+2}=-\frac{1}{4}$