Problem 115 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n}\) is equal to (A) \(-1\) (B) \(e^{-1}\) (C) 1 (D) \(e\)

Step-by-Step Solution

Answer

The limit evaluates to \( e^{-1} \), so the correct answer is (B) \( e^{-1} \).

Step 1: Understand the Limit Expression

The given expression is \( \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \). We need to find the value of this limit as \( n \) approaches infinity.

Step 2: Simplify using Stirling's Approximation

Stirling's approximation for \( n! \) is \( n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \). Applying this, we have:\[(n!)^{1/n} \approx \left( \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \right)^{1/n} = \sqrt[2n]{2 \pi n} \cdot \frac{n}{e}\].

Step 3: Calculate the Expression Inside the Limit

Substitute the approximated \( (n!)^{1/n} \) into the limit expression:\[\lim_{n \to \infty} \frac{\sqrt[2n]{2 \pi n} \cdot \frac{n}{e}}{n} = \lim_{n \to \infty} \frac{n}{e\cdot n} \cdot \sqrt[2n]{2 \pi n}\].

Step 4: Simplify the Limit Expression

Simplify \( \frac{n}{e\cdot n} \) to \( \frac{1}{e} \). Now evaluate the limit:\[\lim_{n \to \infty} \frac{1}{e} \cdot \sqrt[2n]{2 \pi n} \]Since \( \sqrt[2n]{2 \pi n} \to 1 \) as \( n \to \infty \), the limit becomes \( \frac{1}{e} \cdot 1 = \frac{1}{e} \).

Step 5: Interpret the Result

The limit evaluates to \( \frac{1}{e} \) which matches with the option \( e^{-1} \). Therefore, the correct answer is \( e^{-1} \).

Key Concepts

Stirling's approximation

Stirling's approximation is a valuable tool in mathematics when dealing with factorials, especially for large numbers. It's an approximation formula for approximating `n!` which helps simplify complex expressions. The formula is:

• \[ n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \]

Where:

• \( n! \) is the factorial of \( n \), which is the product of all positive integers up to \( n \).
• \( \pi \) is the mathematical constant Pi, approximately equal to 3.14159.
• \( e \) is Euler's number, approximately equal to 2.71828.

Stirling's approximation is valuable for calculating factorials of large numbers because it provides a more manageable expression compared to multiplying a sequence of increasingly large numbers. In the context of evaluating limits, using Stirling's approximation helps see how rapidly factorial expressions grow and how they behave as they approach infinity.

Infinite limits in calculus

In calculus, understanding infinite limits is crucial, especially when analyzing functions or sequences as they stretch towards infinity. An infinite limit usually signifies that as one variable grows larger without bound, the value of the function approaches a particular number.In the provided exercise, the goal was to evaluate:

• \[ \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \]

As \( n \) grows infinitely, the behavior of the factorial function becomes central to solving the problem. Infinite limits often involve approximations, like Stirling's, to simplify the computation of such complex expressions. The limit in the exercise uses these ideas to find that as \( n \) becomes very large, the expression settles to \( \frac{1}{e} \). Mastery of infinite limits in calculus helps in understanding behaviors of competing growth rates, such as polynomials and factorials.

Factorial growth rate

Factorial growth rate is an essential concept when studying the behavior of sequences and functions, especially in combinatorics and calculus. The factorial of a number, `n!`, grows extremely fast since each term in the product is larger than the previous one.Here's an overview of how factorial growth compares to other functions:

• Exponential Growth: Growth pattern follows \( a^n \) where \( a > 1 \), usually dominates polynomial and linear growth but not factorial growth.
• Factorial Growth: Even faster than exponential, \( n! \) grows rapidly such that \( n! > a^n \) for sufficiently large \( n \).

Studying the growth rates is beneficial for estimating efficiencies in algorithms or predicting computational difficulty. In the specific problem, evaluating \((n!)^{1/n}\) helps gain insight into how factorials behave when standardized by the nth root, revealing the relative contribution and effect of each multiplicand in the factorial. Using Stirling's approximation gives us an analytic tool to handle these faster growing sequences and functions.