Question

Como eu faço para calcular esse limite sem utilizar a regra de L'hospital

$\lim_{x \to 1} (\sqrt[2]{x} - 1 ) / ( \sqrt[3]{x} - 1)$

Grato

Answer 1

$\underset{x\to1}{\lim}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}=\underset{x\to1}{\lim}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+1}=$ $\underset{x\to1}{\lim}\frac{x-1}{(\sqrt[3]{x}-1)(\sqrt{x}+1)}\cdot\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\underset{x\to1}{\lim}\frac{(x-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(x-1)(\sqrt{x}+1)}=$ $\underset{x\to1}{\lim}\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\sqrt{x}+1)}=\underset{x\to1}{\lim}\frac{(\sqrt[3]{1^2}+\sqrt[3]{1}+1)}{(\sqrt{1}+1)}=\frac32$

Obs: l'Hopital, sem o s.