## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{h\to0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{-2}{x^3}$
*Notes: In the question that seemingly contains 2 or more variables, you only need to care about the variable mentioned under $\lim$. $\lim\limits_{h\to0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$ $=\lim\limits_{h\to0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$ $=\lim\limits_{h\to0}\frac{x^2-(x+h)^2}{x^2h(x+h)^2}$ $=\lim\limits_{h\to0}\frac{x^2-(x^2+2xh+h^2)}{x^2h(x+h)^2}$ $=\lim\limits_{h\to0}\frac{-2xh-h^2}{x^2h(x+h)^2}$ $=\lim\limits_{h\to0}\frac{-h(2x+h)}{x^2h(x+h)^2}$ $=\lim\limits_{h\to0}\frac{-(2x+h)}{x^2(x+h)^2}$ (both $h$ gets cancelled out) $=\frac{-(2x+0)}{x^2(x+0)^2}$ $=\frac{-2x}{x^4}$ $=\frac{-2}{x^3}$