Problem 114 - 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 is \(e^{-1}\), so the answer is (B).

Step 1: Understand the Limit

We need to evaluate the limit \( \lim_{n \rightarrow \infty} \frac{(n!)^{1/n}}{n} \). This limit involves the factorial of \( n \) and it's raised to the power of \( 1/n \), then divided by \( n \).

Step 2: Use Stirling's Approximation

Stirling's approximation for large \( n \) is \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \). We will use this approximation to simplify \( n! \) and find \( (n!)^{1/n} \).

Step 3: Simplify \((n!)^{1/n}\) Using Stirling's Formula

Using Stirling's approximation, we have \((n!)^{1/n} \approx \left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\right)^{1/n}\). Simplifying, this becomes \( (2 \pi n)^{1/(2n)} \cdot \frac{n}{e} \).

Step 4: Evaluate the Limit of Each Factor

Evaluate the limit of each component as \( n \rightarrow \infty \). For the factor \( (2\pi n)^{1/(2n)} \), note that it approaches 1 as \( n \rightarrow \infty \). This is because the exponent \( \frac{1}{2n} \) goes to 0. The main term \( \frac{n}{e} \) simplifies our problem considerably.

Step 5: Assemble the Limit

Thus, the expression \( \lim_{n \rightarrow \infty} \frac{(n!)^{1/n}}{n} = \lim_{n \rightarrow \infty} \frac{1}{e} = \frac{1}{e} \).

Step 6: Conclusion

Given that \( \lim_{n \rightarrow \infty} \frac{(n!)^{1/n}}{n} = \frac{1}{e} \), the answer is option (B) \( e^{-1} \).

Key Concepts

Factorial

The factorial of a number is a mathematical operation that multiplies a number by every positive integer less than itself. It's denoted by the symbol "!".
The factorial of a non-negative integer, say \( n \), is defined as the product of all positive integers less than or equal to \( n \). This can be mathematically expressed as:

• \( n! = n \, \times \, (n-1) \, \times \, (n-2) \, \times \, ... \, \times \, 1 \)

For example, \( 5! = 5 \, \times \, 4 \, \times \, 3 \, \times \, 2 \, \times \, 1 = 120 \).
The factorial operation grows very quickly as \( n \) increases, making calculations with large numbers challenging without using approximations or computational tools. An important thing to remember is that \( 0! \) is defined as 1.
Understanding factorials is crucial in permutations, combinations, and many other areas of mathematics.

Stirling's approximation

As numbers get larger, computing factorials directly becomes impractical due to their rapid growth. This is where Stirling’s approximation becomes valuable.

Stirling's approximation provides a way to approximate factorials for large \( n \). The approximation is given by:

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

Here, \( e \) represents Euler's number, approximately equal to 2.71828. This formula helps reduce the complexity of math problems that involve large factorials.
Stirling's approximation offers insights into the behavior of factorials and assists in evaluating limits, like our exercise. By substituting this approximation into complex factorial expressions, calculations are made more manageable.
Using Stirling’s approximation helps break down and understand problems where direct calculation of large factorials is not feasible.

Asymptotic Behavior

In mathematics, understanding the asymptotic behavior of functions is vital when examining limits, especially as a variable approaches infinity.
Asymptotic behavior gives insight into how functions behave as their arguments grow very large. For instance, it helps identify terms that dominate a function's growth and those that become negligible.

In the example of the limit \( \lim_{n \rightarrow \infty} \frac{(n!)^{1/n}}{n} \), asymptotic analysis enables us to understand which parts of Stirling's approximation are essential in determining the limit.
Here, simplifying \( (n!)^{1/n} \) using

• \( (2 \pi n)^{1/(2n)} \, \times \, \frac{n}{e} \)

asymptotically approaches \( \frac{1}{e} \) as \( n \) goes to infinity. The term \((2\pi n)^{1/(2n)}\) becomes negligible, resulting in us focusing mainly on \( \frac{n}{e} \).
Asymptotic behavior is an essential tool for solving many real-world problems involving very large numbers or extreme conditions.