Question

A resposta é: 1/3
Preciso do calculo passo a passo pra tirar duvidas...$\lim_{x \to 0} \frac{\sqrt[3]{x+1}-1 }{x}$

Answer 1

Temos que

$\displaystyle\lim_{x \rightarrow 0 } \dfrac{ \sqrt[3]{x + 1} - 1 }{x} \\ = \displaystyle\lim_{x \rightarrow 0 } \dfrac{ \sqrt[3]{x + 1} - 1 }{x} \cdot \dfrac{(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1) }{(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1)} \\ = \displaystyle\lim_{x \rightarrow 0 }\dfrac{( \sqrt[3]{x + 1} - 1) (( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1) }{x(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1)} \\ = \displaystyle\lim_{x \rightarrow 0 }\dfrac{ {(\sqrt[3]{x + 1}) }^{ 3} - {1}^{3} }{x(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1)} \\ = \displaystyle\lim_{x \rightarrow 0 }\dfrac{ x - 1 - 1 }{x(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1)} \\ = \displaystyle\lim_{x \rightarrow 0 }\dfrac{ x }{x(( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1)} \\ = \displaystyle\lim_{x \rightarrow 0 }\dfrac{ 1 }{( \sqrt[3]{x + 1} ) ^{2} + \sqrt[3]{x + 1} + 1} \\ = \dfrac{1}{( \sqrt[3]{0 + 1} ) ^{2} + \sqrt[3]{0+ 1} + 1} \\ = \dfrac{1}{( \sqrt[3]{1})^{2} + \sqrt[3]{1} + 1 } = \dfrac{1}{1 + 1 + 1} = \dfrac{1}{3}$

Eu só usei a diferença de cubos que é dada por

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

Answer 2

Resposta:

Explicação passo-a-passo:

a³ - b³ = (a - b)(a² + ab + b³)

$\lim_{x \to \00} \frac{\sqrt[3]{x+1}-1}{x} = \lim_{t \to \11} \frac{\sqrt[3]{t^{3}}-1}{t^{3}-1}=\lim_{x \to \11}\frac{t-1}{(t-1)(t^{2}+t+1)}= \lim_{t \to \11}\frac{1}{t^{2}+t+1}=\frac{1}{1+1+1}=\frac{1}{3}$

Cálculos auxiliares.

Fazendo x + 1 = t³ ⇒ x = t³ - 1

t³ - 1 = (t - 1)(t² + t + 1)

x → 0 ⇒ t → 1