Problem 42 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} \frac{1}{n}\left(1+e^{1 / n}+e^{2 / n}+\ldots+e^{\frac{n-1}{n}}\right)\) is equal to (A) \(e\) (B) \(-e\) (C) \(e-1\) (D) \(1-e\)

Step-by-Step Solution

Answer

The limit is equal to \( e-1 \), which is option (C).

Step 1: Understand the Expression

The given limit is \( \lim _{n \rightarrow \infty} \frac{1}{n}\left(1+e^{1 / n}+e^{2 / n}+\ldots+e^{(n-1)/n}\right) \). This can be interpreted as a Riemann sum, which is a discrete way to approximate the value of an integral.

Step 2: Recognize the Riemann Sum

A Riemann sum of the form \( \frac{1}{n} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) \) approximates the integral \( \int_{0}^{1} f(x) \, dx \). In this case, \( f\left(\frac{k}{n}\right) = e^{k/n} \).

Step 3: Set Up the Integral

The limit can be recast as an integral \( \int_{0}^{1} e^x \, dx \) because \( \frac{1}{n} \sum_{k=0}^{n-1} e^{k/n} \) approximates this integral as \( n \to \infty \).

Step 4: Calculate the Integral

The integral \( \int_{0}^{1} e^x \, dx \) is calculated as follows: \( \left[ e^x \right]_{0}^{1} = e^1 - e^0 = e - 1 \).

Step 5: Compare with Options

After calculating the integral, we find the result \( e - 1 \), which corresponds to option (C).

Key Concepts

Riemann Sum

A Riemann sum is a fundamental concept in calculus. It allows us to approximate the value of an integral by breaking down a region into smaller slices. Each slice resembles a rectangle. By adding up the areas of these rectangles, we approximate the area under a curve. This is especially helpful for functions that are difficult to integrate outright. In this specific exercise, the expression resembles a Riemann sum of the form \( \frac{1}{n} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) \). If compared to the Riemann sum definition, the role of the function \( f(x) \) is filled by \( e^{x} \). Each term \( e^{k/n} \) acts as the height of a rectangle, while \( \frac{1}{n} \) is the width. This setup approximates the integral \( \int_{0}^{1} e^x \, dx \) as \( n \) approaches infinity.With more terms included (or as \( n \to \infty \)), the approximation becomes more precise, providing a clearer estimation of the true integral.

Integral Calculus

Integral calculus deals with finding the total size or value, such as areas under curves. Integrals are the inverse operation of derivatives in calculus, allowing us to sum continuous functions over intervals. The exercise uses a limit expression to transition from a Riemann sum to an integral. While Riemann sums use discrete intervals to approximate a function, integrals give the exact value over a continuous interval. The specific integral in this scenario, \( \int_{0}^{1} e^x \, dx \), involves the exponential function. Solving it involves finding an antiderivative, which means identifying a function whose derivative matches the function inside the integral. To calculate this, the exponential function retains its form, leading to:

• \( \int e^x \, dx = e^x + C \)

where \( C \) is the constant of integration (used in indefinite integrals). For a definite integral from 0 to 1, we evaluate this function at the upper and lower limits, resulting in \( [e^x]_{0}^{1} = e^1 - e^0 = e - 1 \). This precise solution helps confirm the answer \( e - 1 \) for the problem.

Exponential Function

Exponential functions are a vital part of many areas in mathematics. The function \( e^x \) specifically is frequently seen due to its unique properties. An exponential function is defined as \( f(x) = e^x \), where \( e \) (approximately 2.718) is the base of natural logarithms. Its special property is that its rate of change is proportional to its current value. In the given exercise, \( e^{x} \) appears as part of the Riemann sum and integral. Each term, like \( e^{k/n} \), represents values of the function at specific points. These exponential terms are part of the summation process used to approximate the integral.The reason \( e^{x} \) is so ubiquitous is due to these key features:

• Its derivative is exactly itself \( \frac{d}{dx}e^x = e^x \)
• The function never crosses the x-axis—all values are positive

This exercise effectively uses these properties to transition from a sum to an integral, showcasing the simplicity and elegance of the exponential function in calculus.