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Chapter 1: Problem 71

The value of $\lim _{n \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}}$ is equal to (A) \(]\) (B) 0 (C) \(\left(\frac{1}{\mathrm{e}}\right)^{2 / 5}\) (D) \(\mathrm{e}^{2 / 3}\)

Short Answer

Expert verified
Answer: (A) 1

Step by step solution

01

Rewrite the given expression using logarithms

Since we are dealing with exponentials, it would be easier to work with the logarithms and then convert back later. Let's rewrite the given expression in logarithmic form: Let L = \(\lim _{n \rightarrow \infty}\left(\frac{n!}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}}\). Then \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(\frac{n!}{n^n}\right)\)
02

Apply Stirling's Approximation

Stirling's approximation states that for large values of n, \(n! \approx n^ne^{-n}\sqrt{2\pi n}\). Now, let's substitute this approximation in the logarithmic expression: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(\frac{n^ne^{-n}\sqrt{2\pi n}}{n^n}\right)\)
03

Simplify the expression

Now simplify the expression inside the logarithm and factor out constants: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(e^{-n}\sqrt{2\pi n}\right) = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \left(-n + \frac{1}{2}\ln(2\pi n) \right)\)
04

Apply L'H么pital's rule

Since we have the indeterminate form of the type \(\frac{\infty}{\infty}\), let's apply L'H么pital's rule by taking derivatives of the numerator and denominator with respect to n: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{-1 + \frac{1}{2n}}{10n} = \lim _{n \rightarrow \infty} \frac{\frac{1}{2n} - 1}{10n}\) Now applying L'H么pital's rule again: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{-\frac{1}{2n^2}}{10} = 0\)
05

Calculate the final result

Since \(\ln{L} = 0\), we can find the value of L by taking the exponent with the base e: \(L = e^0 = 1\) So, the correct answer is: \(\lim _{n \rightarrow \infty}\left(\frac{n!}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}} = 1\) which corresponds to the option (A).

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Most popular questions from this chapter

Assertion (A): $\lim _{x \rightarrow 0}[x]\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$ where [.] represents greatest integer function does not exist. Reason $(\mathrm{R}): \lim _{\mathrm{x} \rightarrow 0}\left(\frac{\mathrm{e}^{\mathrm{l} / \mathrm{x}}-1}{\mathrm{e}^{1 / \mathrm{x}}+1}\right)$ does not exist.

If \(\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x-b\right)=0\), then for \(k \geq 2, k \in\) \(\mathrm{N}\) which of the following is/are correct ? (A) \(2 \mathrm{a}+\mathrm{b}=0\) (B) \(a+2 b=0\) (C) \(\lim _{n \rightarrow \infty} \sec ^{2 n}(k ! \pi b)=1\) (D) \(\lim _{n \rightarrow \infty} \sec ^{2 n}(k ! \pi a)=1\)

The value of $\left(\lim _{x \rightarrow 0}\left[\frac{100 x}{\sin x}\right]+\left[\frac{99 \sin x}{x}\right]\right)$ is (where [.] denotes greatest integer function) (A) 199 (B) 198 (C) 197 (D) None of these

$\lim _{n \rightarrow \infty} \frac{1 . n+(n-1)(1+2)+(n-2)(1+2+3)+. .1 \cdot \sum_{r=1}^{n} r}{n^{4}}$ is equal to (A) \(1 / 12\) (B) \(1 / 24\) (C) \(1 / 6\) (D) \(1 / 48\)

If \(\mathrm{k}\) is an integer such that $\lim _{n \rightarrow \infty}\left(\left(\cos \frac{k \pi}{4}\right)^{n}-\left(\cos \frac{k \pi}{6}\right)^{n}\right)=0$, then (A) \(\mathrm{k}\) is divisible neither by 4 nor by 6 (B) \(\mathrm{k}\) must be divisible by 12 , but not necessarily by 24 (C) \(\mathrm{k}\) must be divisible by 24 (D) either \(\mathrm{k}\) is divisible by 24 or \(\mathrm{k}\) is divisible neither by 4 nor by 6
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