Problem 76 - Calculus for JEE ADVANCED

Question

\(L=\lim _{n \rightarrow \infty} \frac{e^{n}}{\left(1+\frac{1}{n}\right)^{n^{2}}}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Limit Expression

We are tasked with finding the limit \( L = \lim_{n \rightarrow \infty} \frac{e^n}{\left(1+\frac{1}{n}\right)^{n^2}} \). This expression involves an exponential function and a base raised to a quadratic power.

Step 2: Simplify Using Known Limits

Recall the limit: \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \). We can use this to recognize a similar structure. Therefore, express the denominator as \( \left(1 + \frac{1}{n}\right)^{n^2} = \left(\left(1 + \frac{1}{n}\right)^n\right)^{n} \approx e^n^{n} = e^n \).

Step 3: Calculate the Denominator's Limit Exponentially

Further, observe the expression \( \left(1 + \frac{1}{n}\right)^{n^2} \). Using the expansion of the expression with binomial theorem and Taylor's expansion, it approaches \( e^n \) as \( n \to \infty \).

Step 4: Evaluate the Overall Limit

Since the numerator is \( e^n \) and the denominator approaches \( e^n \) as \( n \to \infty \), the overall expression becomes \( \frac{e^n}{e^n} \). This simplifies directly to 1 when evaluated at the limit of \( n \to \infty \).

Step 5: Conclusion

As \( n \to \infty \), the limit \( L = \frac{e^n}{\left(1+\frac{1}{n}\right)^{n^2}} \) simplifies to 1. Therefore, \( L = 1 \).

Key Concepts

Exponential Functions

Exponential functions are central to calculus and appear frequently when analyzing limits and growth patterns. In mathematics, an exponential function has the form \( f(x) = a^x \) where \( a \) is a positive constant. These functions are widely recognized for their rapid growth. For example, \( e^x \) is the standard exponential function with base \( e \), which is approximately equal to 2.71828.Exponential growth is characterized by:

• The function’s rate of growth is proportional to its size.
• The function's derivative \( f'(x) = a^x \ln a \) showcases a directly proportional relationship with the original function.
• Common occurrences include compound interest and population growth.

In our original problem, \( e^n \) is the exponential function in the numerator. Understanding the behavior of this function is key to solving limits like the one provided in the exercise. Complexities of limits involving exponential functions often require simplifying expressions using known limit properties.

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra, used to expand expressions raised to a power. It states that:\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]This formula lets you simplify powers that look complicated. The terms \( \binom{n}{k} \) represent binomial coefficients and describe the number of ways to choose \( k \) elements from \( n \) without regard to order.Why is the Binomial Theorem useful? Here are some contexts:

• It provides an expansion of any power of a binomial expression.
• It serves in approximating roots and powers of numbers.
• In calculus, it's seen in limit problems and series expansions.

In the exercise, its conceptual application helps us understand how \( \left(1 + \frac{1}{n}\right)^{n^2} \) behaves as \( n \to \infty \). Recognizing similar structures to those in the theorem aids in recognizing that the expression approaches a well-known limit.

Taylor Series Expansion

The Taylor series expansion is another powerful mathematical concept that helps us approximate functions around a certain point using polynomial expressions. The formula for a Taylor series centered at \( a \) is:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]This allows for expressing complex functions as infinite series of polynomials. When people encounter functions, like exponentials, the Taylor series can approximate these functions:

• Making it easier to calculate limits, derivatives, and integrals.
• Giving insight into function behavior near a specific point.
• Offering solutions to differential equations.

In our given exercise, the Taylor Series provides a frame to understand the expansion of \( \left(1 + \frac{1}{n}\right)^n \). It shows why as \( n \to \infty \), the expression converges to \( e \), particularly in complex limit evaluations. Taylor expansions are crucial in demystifying how elements of an expression interact at infinity.