JEE PYQ: Limits Question 13
Question 13 - 2020 (03 Sep Shift 2)
$\lim_{x \to a} \frac{(a + 2x)^{1/3} - (3x)^{1/3}}{(3a + x)^{1/3} - (4x)^{1/3}}$ $(a \neq 0)$ is equal to:
(1) $\left(\frac{2}{3}\right)^{4/3}$
(2) $\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{1/3}$
(3) $\left(\frac{2}{9}\right)^{4/3}$
(4) $\left(\frac{2}{9}\right)\left(\frac{2}{9}\right)^{1/3}$
Show Answer
Answer: (2)
Solution
Using L’Hospital rule: $\lim_{x\to a} \frac{\frac{1}{3}(a+2x)^{-2/3}\cdot 2 - \frac{1}{3}(3x)^{-2/3}\cdot 3}{\frac{1}{3}(3a+x)^{-2/3}\cdot 1 - \frac{1}{3}(4x)^{-2/3}\cdot 4} = \frac{\frac{1}{3}(3a)^{-2/3}(2-3)}{\frac{1}{3}(4a)^{-2/3}(1-4)} = \frac{2^{4/3}}{9^{1/3}} \cdot \frac{1}{3} = \frac{2}{3}\left(\frac{2}{9}\right)^{1/3}$.
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