First slide

Introduction to limits

Question

$lim_{x \to a} \frac{(a + 2 x)^{1 / 3} - (3 x)^{1 / 3}}{(3 a + x)^{1 / 3} - (4 x)^{1 / 3}}, (a \neq 0)$ is equal to

Moderate

A

$\frac{2}{9} {(\frac{2}{3})}^{1 / 3}$
B

${(\frac{2}{3})}^{1 / 3}$
C

${(\frac{2}{9})}^{4 / 3}$
D

$\frac{2}{3} {(\frac{2}{9})}^{1 / 3}$

Solution

Let $L = lim_{x \to a} \frac{(a + 2 x)^{1 / 3} - (3 x)^{1 / 3}}{(3 a + x)^{1 / 3} - (4 x)^{1 / 3}} .$ Applying $L'$
Hospital's rule, we obtain
$\begin{array}{l} L = lim_{x \to a} \frac{\frac{2}{3} (a + 2 x)^{- 2 / 3} - (3 x)^{- 2 / 3}}{\frac{1}{3} (3 a + x)^{- 2 / 3} - \frac{4}{3} (4 x)^{- 2 / 3}} 1 \\ \Rightarrow L = \frac{- \frac{1}{3} (3 a)^{- 2 / 3}}{- (4 a)^{- 2 / 3}} \\ \Rightarrow L = \frac{1}{3} \frac{(4 a)^{2 / 3}}{(3 a)^{2 / 3}} = \frac{2^{4 / 3}}{3^{5 / 3}} = \frac{2}{3} {(\frac{2}{9})}^{1 / 3} \end{array}$

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