Answer
$\lim_{x\to-4}\dfrac{\dfrac{1}{4}+\dfrac{1}{x}}{4+x}=-\dfrac{1}{16}$
Work Step by Step
$\lim_{x\to-4}\dfrac{\dfrac{1}{4}+\dfrac{1}{x}}{4+x}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to-4}\dfrac{\dfrac{1}{4}+\dfrac{1}{x}}{4+x}=\dfrac{\dfrac{1}{4}+\dfrac{1}{-4}}{4-4}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Evaluate the sum of fractions in the numerator and simplify:
$\lim_{x\to-4}\dfrac{\dfrac{1}{4}+\dfrac{1}{x}}{4+x}=\lim_{x\to-4}\dfrac{\dfrac{4+x}{4x}}{4+x}=\lim_{x\to-4}\dfrac{4+x}{4x(4+x)}=...$
$...=\lim_{x\to-4}\dfrac{1}{4x}=...$
Try to evaluate the limit using direct substitution again:
$...=\lim_{x\to-4}\dfrac{1}{4x}=\dfrac{1}{4(-4)}=-\dfrac{1}{16}$
