Question

If the value of $\underset{x\to -2}{lim}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$ is $k$, then find the value of $4k$

• A. $-\frac{1}{4}$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

The correct option is C $-1$

$k=\underset{x\to -2}{lim}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\to \frac{0}{0}$form
Apply L'Hospital's rule
$k=\underset{x\to -2}{lim}\frac{-\frac{1}{x^2}}{1}=-\frac{1}{x^2}=-\frac{1}{4}$
$\therefore 4k=4\times -\frac{1}{4}=-1$