First slide

Introduction to limits

Question

The value of $lim_{x \to \infty} (\sqrt[3]{x^{3} + x^{2}} - \sqrt[3]{x^{3} - x^{2}}),$ is

Moderate

A

$\frac{1}{3}$
B

$\frac{2}{3}$
C

1
D

$\frac{4}{3}$

Solution

We know that $a - b = \frac{a^{3} - b^{3}}{a^{2} + a b + b^{2}}$
$∴ lim_{x \to \infty} (\sqrt[3]{x^{3} + x^{2}} - \sqrt[3]{x^{3} - x^{2}})$
$\begin{array}{l} = lim_{x \to \infty} \frac{(x^{3} + x^{2}) - (x^{3} - x^{2})}{{(x^{3} + x^{2})}^{2 / 3} + {(x^{3} - x^{2})}^{1 / 3} {(x^{3} + x^{2})}^{1 / 3} + {(x^{3} - x^{2})}^{2 / 3}} \\ = lim_{x \to \infty} \frac{2 x^{2}}{{(x^{3} + x^{2})}^{2 / 3} + {(x^{3} - x^{2})}^{1 / 3} {(x^{3} + x^{2})}^{1 / 3} + {(x^{3} - x^{2})}^{2 / 3}} \\ = lim_{x \to \infty} \frac{2}{{(1 + \frac{1}{x})}^{2 / 3} + {(1 - \frac{1}{x})}^{1 / 3} {(1 + \frac{1}{x})}^{1 / 3} + {(1 - \frac{1}{x})}^{2 / 3}} \end{array}$
$= \frac{2}{1 + 1 + 1} = \frac{2}{3}$

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