Question

$\underset{x\to -2}{lim}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

Answer 1

We have,

${lim}_{x\to -2}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

${lim}_{x\to -2}\frac{x+2}{2x(x+2)}$

${lim}_{x\to -2}\frac{1}{2x}$

$=\frac{1}{2\times -2}$

$=-\frac{1}{4}$