## Thomas' Calculus 13th Edition

$-\dfrac{1}{4}$
Here, we have $f(x)=\frac{1}{x};x=-2$ and $f(x+h)=\dfrac{1}{x+h}$ Now, $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{(\dfrac{1}{x+h})-(\dfrac{1}{x})}{h}=\lim\limits_{h\to0}\dfrac{-1}{x(x+h)}$ Plug $x=-2$, we get $\lim\limits_{h\to0}\dfrac{-1}{x(x+h)}=\dfrac{-1}{(-2)(-2+0)}$ or, $\dfrac{-1}{(-2)(-2)}=-\dfrac{1}{4}$