JEE PYQ: Limits Question 25
Question 25 - 2019 (10 Apr Shift 1)
$\lim_{n \to \infty} \frac{(n+1)^{1/3} + (n+2)^{1/3} + \cdots + (2n)^{1/3}}{n^{4/3}}$ is equal to:
(1) $\frac{3}{4}(2)^{4/3} - \frac{3}{4}$
(2) $\frac{4}{3}(2)^{4/3}$
(3) $\frac{3}{2}(2)^{4/3} - \frac{3}{4}$
(4) $\frac{3}{4}(2)^{3/4}$
Show Answer
Answer: (1)
Solution
$= \lim_{n\to\infty}\sum_{r=1}^{n}\frac{(n+r)^{1/3}}{n^{4/3}} = \int_0^1 (1+x)^{1/3},dx = \left[\frac{3}{4}(1+x)^{4/3}\right]_0^1 = \frac{3}{4}(2)^{4/3} - \frac{3}{4}$.
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