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