Question

Calcule o seguinte limite lim h->0 ((1/(x+h)^2)-(1/x^2))/h
Obs: Achei a resposta usando L'Hopital, queria saber como eliminar a indeterminação sem usar L'Hopital.. Segue a foto da questão!

Answer 1

$\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{x^{2}+2xh+h^{2}}-\frac{1}{x^{2}}}{h}$

Fazendo a subtração de frações no numerador:

$\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{\frac{x^{2}-(x^{2}+2xh+h^{2})}{x^{2}(x^{2}+2xh+h^{2})}}{h}\\\\\\\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{x^{2}-x^{2}-2xh-h^{2}}{hx^{2}(x^{2}+2xh+h^{2})}\\\\\\\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{-2xh-h^{2}}{hx^{2}(x^{2}+2xh+h^{2})}$

Colocando h em evidência no numerador:

$\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{h(-2x-h)}{hx^{2}(x^{2}+2xh+h^{2})}\\\\\\\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\lim\limits_{h\rightarrow0}~\dfrac{-2x-h}{x^{2}(x^{2}+2xh+h^{2})}\\\\\\\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\dfrac{-2x-0}{x^{2}(x^{2}+2x(0)+0^{2})}\\\\\\\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=\dfrac{-2x}{x^{4}}$

Simplificando:

$\boxed{\boxed{\lim\limits_{h\rightarrow0}~\dfrac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}=-\dfrac{2}{x^{3}}}}$