Question

Questão i, com resoluçao por favor

Answer 1

$\lim_{h\rightarrow0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\\\\\\\lim_{h\rightarrow0}\frac{\left(\frac{1}{(x+h)}+\frac{1}{x}\right)\left(\frac{1}{(x+h)}-\frac{1}{x}\right)}{h}=\\\\\\\lim_{h\rightarrow0}\frac{\left(\frac{x+x+h}{x(x+h)}\right)\left(\frac{x-x-h}{x(x+h)}\right)}{h}=$

$\lim_{h\rightarrow0}\frac{\frac{2x+h}{x(x+h)}\cdot\left(\frac{-h}{x(x+h)}\right)}{h}=\\\\\\\lim_{h\rightarrow0}\frac{2x+h}{x(x+h)}\cdot\frac{-h}{x(x+h)}\div h=$

$\lim_{h\rightarrow0}\frac{2x+h}{x(x+h)}\cdot\frac{-h}{x(x+h)}\cdot\frac{1}{h}= \\\\\\\lim_{h\rightarrow0}\frac{2x+h}{x(x+h)}\cdot\frac{-1}{x(x+h)}=\\\\\\\lim_{h\rightarrow0}-\frac{2x+h}{x^2(x+h)^2}=\\\\\\-\frac{2x+0}{x^2(x+0)^2}=\\\\\\-\frac{2x}{x^4}=\\\\\\\boxed{-\frac{2}{x^3}}<span>$

Ou, $\boxed{-2x^{-3}}$