Given,

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$

Substituting $x =  – 2$ to get,

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}} = \frac{0}{0}$

As, this limit is undefined so the other approach would be,

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}} = \frac{{\frac{{2 + x}}{{2x}}}}{{x + 2}}$

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}} = \frac{1}{{2x}}$

Now, substituting $x =  – 2$ to get,

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}} = \frac{1}{{2( – 2)}}$

$\mathop {\lim }\limits_{x \to  – 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}} =  – \frac{1}{4}$.