Question

Quero o calculo a resposta é; -3/8

$\lim_{x \to -1} \frac{x+1}{\sqrt[3]{2x+3} -1}$

Answer 1

Na verdade a resposta é 3/2. Usando a igualdade

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

onde

$a = \sqrt[3]{2x + 3} \: \: \: \: e \: \: \: \: b = 1$

Temos

$\displaystyle\lim_{x \to -1} \frac{x+1}{\sqrt[3]{2x+3} -1} \\ = \displaystyle\lim_{x \to -1} \frac{x+1}{\sqrt[3]{2x+3} -1} \cdot\frac{((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1)}{((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1)} \\ =\displaystyle\lim_{x \to -1} \frac{(x+1)((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1) }{(\sqrt[3]{2x+3} )^{3} - {1}^{3} } \\ = \displaystyle\lim_{x \to -1} \frac{(x+1)((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1) }{2x+3 - 1 } \\ = \displaystyle\lim_{x \to -1} \frac{(x+1)((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1) }{2x+2 } \\ = \displaystyle\lim_{x \to -1} \frac{(x+1)((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1) }{2(x + 1) } \\ = \displaystyle\lim_{x \to -1} \frac{((\sqrt[3]{2x+3})^{2} + \sqrt[3]{2x + 3}+1) }{2} \\ = \dfrac{((\sqrt[3]{1})^{2} + \sqrt[3]{1}+1) }{2} = \frac{1 + 1 + 1}{2} = \frac{3}{2}$