JEE PYQ: Limits Question 6
Question 6 - 2021 (25 Feb Shift 1)
$\lim_{n \to \infty} \left(1 + \frac{1 + \frac{1}{2} + \cdots + \frac{1}{n}}{n^2}\right)^n$ is equal to:
(1) $\frac{1}{2}$
(2) $\frac{1}{e}$
(3) $1$
(4) $0$
Show Answer
Answer: (3)
Solution
It is $1^\infty$ form. $L = e^{\lim_{n\to\infty} \frac{S}{n}}$ where $S = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. Since $S < 1 + 1 + 1 + \cdots + 1$ (bounded by $\ln n + 1$), $\frac{S}{n} \to 0$. So $L = e^0 = 1$.
BETA
