Answer
$\lim\limits_{x\to3}\frac{\frac{1}{x}-\frac{1}{3}}{x-3}=\frac{-1}{9}$
Work Step by Step
$\lim\limits_{x\to3}\frac{\frac{1}{x}-\frac{1}{3}}{x-3}$
$=\lim\limits_{x\to3}\frac{\frac{3-x}{3x}}{x-3}$
$=\lim\limits_{x\to3}\frac{3-x}{3x(x-3)}$
$=\lim\limits_{x\to3}\frac{-(x-3)}{3x(x-3)}$ (since $a-b=-(b-a)$)
$=\lim\limits_{x\to3}\frac{-1}{3x}$ (divide both numerator and denominator by $x-3$)
$=\frac{-1}{3\times3}$
$=\frac{-1}{9}$
