[LIMITE]

por **tatianaCAL** » Ter Jun 25, 2013 22:26

Como calculo $\lim_{x \to \8}\ \frac{1 - \sqrt[3]{x}}{\sqrt{2x} - 4}$ ?

por **young_jedi** » Qua Jun 26, 2013 22:31

temos que

$\lim_{x\to\infty}\frac{1-x^{\frac{1}{3}}}{x^{\frac{1}{2}}\sqrt2-4}$

$\lim_{x\to\infty}\frac{x^{\frac{1}{2}}(\frac{1}{x^{\frac{1}{2}}}-\frac{1}{x^{\frac{1}{6}}})}{x^{\frac{1}{2}}(\sqrt2-\frac{4}{x^{\frac{1}{2}}})}$

$\lim_{x\to\infty}\frac{\frac{1}{x^{\frac{1}{2}}}-\frac{1}{x^{\frac{1}{6}}}}{\sqrt2-\frac{4}{x^{\frac{1}{2}}}}=\frac{0}{\sqrt2}=0$

por **young_jedi** » Qua Jun 26, 2013 22:32

temos que

$\lim_{x\to\infty}\frac{1-x^{\frac{1}{3}}}{x^{\frac{1}{2}}\sqrt2-4}$

$\lim_{x\to\infty}\frac{x^{\frac{1}{2}}(\frac{1}{x^{\frac{1}{2}}}-\frac{1}{x^{\frac{1}{6}}})}{x^{\frac{1}{2}}(\sqrt2-\frac{4}{x^{\frac{1}{2}}})}$

$\lim_{x\to\infty}\frac{\frac{1}{x^{\frac{1}{2}}}-\frac{1}{x^{\frac{1}{6}}}}{\sqrt2-\frac{4}{x^{\frac{1}{2}}}}=\frac{0}{\sqrt2}=0$

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