limx→0 (27+x)1/3−39−(27+x)2/3 is equal - Infinity Learn by Sri Chaitanya

A
-1/3
B
1/6
C
-1/6
D
1/3

Solution:

We have, $lim_{x \to 0} \frac{(27 + x)^{1 / 3} - 3}{9 - (27 + x)^{2 / 3}}$

$\begin{array}{l} = lim_{x \to 0} \frac{3 [{(1 + \frac{x}{27})}^{1 / 3} - 1]}{9 [1 - {(1 + \frac{x}{27})}^{2 / 3}]} = lim_{x \to 0} \frac{[(1 + \frac{x}{3 \times 27} + \dots) - 1]}{3 [1 - (1 + \frac{2 x}{3 \times 27} + \dots)]} \\ = lim_{x \to 0} \frac{1 [x / 81]}{3 [- 2 x / 81]} = - \frac{1}{6} \end{array}$

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