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