Limits

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Limit $n \to \infty$ of $\frac{1}{n} [(n + 1) (n + 2) ... (n + n)]^{1/ n}$ is:

Select the correct option:

$4/ e$

$2/ e$

$e /4$

$e /2$

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About This Question

Subject: mathematics
Chapter: limit, continuity and differentiability
Topic: limits
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$4/ e$

Let $L$ be the limit. Simplify term: $L = lim_{n \to \infty} [\frac{( n + 1 )}{n} \frac{( n + 2 )}{n} ... \frac{( n + n )}{n}]^{1/ n}$ $L = lim \prod_{r = 1}^{n} (1 + r / n)^{1/ n}$ .
Take log: $ln L = lim \frac{1}{n} \sum_{r = 1}^{n} ln (1 + r / n)$ .
Convert to Integral: $ln L = \int_{0}^{1} ln (1 + x) d x$ .
Evaluate Integral: $\int ln (1 + x) d x = [(1 + x) ln (1 + x) - x]_{0}^{1}$ $= (2 ln 2 - 1) - (0 - 0) = ln 4 - 1 = ln 4 - ln e = ln (4/ e)$ .

So $L = 4/ e$ .

This medium difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of limits. It appeared in the 2025 exam.

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Limits

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