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Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 1 · Q32

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Limits Continuity and Differentiability
2021 · 25 Feb · Shift 1 · Q32

Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 1 · Q32

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}

is equal to :

A
${{1 \over 2}}$
B
1
C
0
D
${{1 \over e}}$

View written solutionFree

Correct answer: B

Let $H_n=1+\frac12+\frac13+\cdots+\frac1n$ be the $n$ th harmonic sum.

The given limit is $L=\lim_{n\to\infty}\left(1+\frac{H_n}{n^2}\right)^n.$

We use the standard fact that $H_n\sim \ln n,$ so in particular $H_n$ grows much slower than $n$ . Hence $\frac{H_n}{n^2}\to 0.$
Now rewrite the expression in exponential form using the standard limit idea:

=\exp\left[n\ln\left(1+\frac{H_n}{n^2}\right)\right].$$ Since $\frac{H_n}{n^2}\to 0$, we use $$\ln(1+x)\sim x \quad (x\to 0).$$ Therefore, $$n\ln\left(1+\frac{H_n}{n^2}\right) \sim n\cdot \frac{H_n}{n^2} =\frac{H_n}{n}.$$ 4. But $$\frac{H_n}{n}\to 0$$ because $H_n\sim \ln n$ and $$\frac{\ln n}{n}\to 0.$$ Hence, $$n\ln\left(1+\frac{H_n}{n^2}\right)\to 0.$$ So, $$L=\exp(0)=1.$$ 5. Checking options: - A: $\frac12$ — incorrect - B: $1$ — correct - C: $0$ — incorrect - D: $\frac1e$ — incorrect Thus the required limit is $$\boxed{1}.$$

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