Answer
$\lim\limits_{x\to0}\dfrac{[1/(x+4)]-(1/4)}{x}=-\dfrac{1}{16}.$
Work Step by Step
$f(x)=\dfrac{[1/(x+4)]-(1/4)}{x}=\dfrac{4-(x+4)}{x(4)(x+4)}=-\dfrac{1}{4x+16}=g(x).$
The function $g(x)$ agrees with the function $f(x)$ at all points except $x=0$. Therefore we find the limit as x approaches $0$ of $f(x)$ by substituting the value into $g(x)$.
$\lim\limits_{x\to0}\dfrac{[1/(x+4)]-(1/4)}{x}=\lim\limits_{x\to0}\dfrac{-1}{4x+16}=-\dfrac{1}{4(0)+16}=-\dfrac{1}{16}.$
