Answer
$\lim\limits_{\Delta x\to0}\dfrac{\dfrac{1}{x+\Delta x+3}-\dfrac{1}{x+3}}{\Delta x}=-\dfrac{1}{(x+3)^2}.$
Work Step by Step
$\lim\limits_{\Delta x\to0}\dfrac{\dfrac{1}{x+\Delta x+3}-\dfrac{1}{x+3}}{\Delta x}$
$=\lim\limits_{\Delta x\to0}\dfrac{(x+3)-(x+\Delta x+3)}{\Delta x(x+\Delta x +3)(x+3)}$
$=\lim\limits_{\Delta x\to0}\dfrac{-\Delta x}{\Delta x(x+\Delta x+3)(x+3)}$
$=\lim\limits_{\Delta x\to0}\dfrac{-1}{(x+\Delta x+3)(x+3)}$
$=\dfrac{-1}{(x+0+3)(x+3)}=-\dfrac{1}{(x+3)^2}.$
