Answer
$\lim\limits_{x\to0}\dfrac{[1/(3+x)]-(1/3)}{x}=-\dfrac{1}{9}.$
Work Step by Step
$f(x)=\dfrac{[1/(3+x)]-(1/3)}{x}=\dfrac{3-(3+x)}{x(3)(3+x)}=\dfrac{-1}{3x+9}=g(x).$
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 $g(x)$.
$\lim\limits_{x\to0}\dfrac{[1/(3+x)]-(1/3)}{x}=\lim\limits_{x\to0}\dfrac{-1}{3x+9}=\dfrac{-1}{3(0)+9}=-\dfrac{1}{9}.$
