Seja:
\(\lim _{x\to 0}\left(\frac{\left(1-\sqrt[3]{1-x}\right)}{\left(1+\sqrt[3]{3x-1}\right)}\right)\)
Veja que quando aplicamos \(x=0\), ocorre uma indeterminação do tipo \(0/0\) . Assim, podemos usar L'hospital, derivando em cima e embaixo:
\(\frac{d}{dx}\left(\left(1-\sqrt[3]{1-x}\right)\right)=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(\sqrt[3]{1-x}\right)\)
Mas:
\(\frac{d}{dx}\left(1\right)=0\\ \frac{d}{dx}\left(\sqrt[3]{1-x}\right)=-\frac{1}{3\left(1-x\right)^{\frac{2}{3}}}\)
Assim:
\(\frac{d}{dx}\left(1-\sqrt[3]{1-x}\right)=\frac{1}{3\left(-x+1\right)^{\frac{2}{3}}}\)
Derivando embaixo:
\(\frac{d}{dx}\left(1+\sqrt[3]{3x-1}\right)=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(\sqrt[3]{3x-1}\right)\)
Mas:
\(\frac{d}{dx}\left(1\right)=0\\ \frac{d}{dx}\left(\sqrt[3]{3x-1}\right)=\frac{1}{\left(3x-1\right)^{\frac{2}{3}}}\)
O limite então fica:
\(\lim _{x\to 0}\left(\frac{\left(1-\sqrt[3]{1-x}\right)}{\left(1+\sqrt[3]{3x-1}\right)}\right)=\lim _{x\to 0}\left(\frac{\frac{1}{3\left(-x+1\right)^{\frac{2}{3}}}}{\left(\frac{1}{\left(3x-1\right)^{\frac{2}{3}}}\right)}\right)\)
Aplicando \(x=0\):
\(\lim _{x\to 0}\left(\frac{\frac{1}{3\left(-x+1\right)^{\frac{2}{3}}}}{\left(\frac{1}{\left(3x-1\right)^{\frac{2}{3}}}\right)}\right)=\left(\frac{\frac{1}{3\left(-0+1\right)^{\frac{2}{3}}}}{\left(\frac{1}{\left(3.0-1\right)^{\frac{2}{3}}}\right)}\right)\)
Simplificando:
\(\left(\frac{\frac{1}{3\left(-0+1\right)^{\frac{2}{3}}}}{\left(\frac{1}{\left(3.0-1\right)^{\frac{2}{3}}}\right)}\right)=\left(\frac{\frac{1}{3}}{\left(\frac{1}{1}\right)}\right)=\frac{1}{3}\)
Portanto:
\(\boxed{\lim _{x\to 0}\left(\frac{\left(1-\sqrt[3]{1-x}\right)}{\left(1+\sqrt[3]{3x-1}\right)}\right)=\frac{1}{3}}\)
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