Problem 7 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{1 / n}\) is equal to (A) \(\frac{2}{-}\) (B) \(\frac{e}{2}\) (C) \(\frac{e}{4}\) (D) \(\frac{4}{e}\)

Step-by-Step Solution

Answer

The limit is \( \frac{e}{4} \), corresponding to option (C).

Step 1: Analyze the Expression Inside the Limit

We need to evaluate the expression \( \left(1 + \frac{1}{n}\right)\left(1 + \frac{2}{n}\right) \cdots \left(1 + \frac{n}{n}\right) \). Notice this can be rewritten as the product \( \prod_{k=1}^{n} \left(1 + \frac{k}{n}\right) \).

Step 2: Use the Logarithm for Simplification

We know the limit of a product can be easier to handle if expressed in logarithmic form. Consider \( \ln \left(\prod_{k=1}^{n} \left(1 + \frac{k}{n}\right)\right) \). The logarithm of a product is the sum of the logarithms: \( \sum_{k=1}^{n} \ln\left(1 + \frac{k}{n}\right) \).

Step 3: Simplify Using an Integral Approximation

We approximate \( \sum_{k=1}^{n} \ln\left(1 + \frac{k}{n}\right) \) as a Riemann sum for the function \( \ln(1+x) \) over the interval \([0, 1]\). As \( n \to \infty \), this sum resembles the integral \( \int_0^1 \ln(1+x) \; dx \).

Step 4: Compute the Integral

To solve \( \int_0^1 \ln(1+x) \; dx \), we use integration by parts:Let \( u = \ln(1+x) \) and \( dv = dx \). Then \( du = \frac{1}{1+x}dx \) and \( v = x \). The integral becomes:\[\int \ln(1+x) \; dx = x \ln(1+x) - \int \frac{x}{1+x} \; dx\] The second integral simplifies to \( x - \ln(1+x) \) by splitting the fraction and integrating term by term. Evaluating from 0 to 1, the integral results in \( \left.\left[x\ln(1+x) - x + \ln(1+x)\right]\right|_0^1 = \frac{1}{2} - \ln(2) \).

Step 5: Evaluate the Original Limit Using the Integral Result

With the integral evaluated, recall the product in the original limit is raised to the power \( \frac{1}{n} \). Therefore, the original limit becomes:\[\lim_{n \to \infty} \exp\left(\frac{1}{n} \cdot \sum_{k=1}^{n} \ln\left(1 + \frac{k}{n}\right)\right) = \exp\left(\int_0^1 \ln(1+x) \; dx \right) = \exp\left(\frac{1}{2} - \ln(2)\right)\]which simplifies to \( \frac{e^{1/2}}{2} = \frac{e}{4} \).

Step 6: Conclusion: Compare with Given Options

The established result \( \frac{e}{4} \) matches option (C). Thus, the limit is \( \frac{e}{4} \).

Key Concepts

Riemann Sum

In calculus, the Riemann sum is a method for approximating the total area under a curve on a graph, otherwise known as the integral. To approximate an integral, you divide the area into small segments and add up the value of the function at specific points within those segments. This technique approaches the actual integral as the number of segments increases.

Notably, as you increase the number of subdivisions (denoted by \( n \)), the approximation improves, ultimately converging to the exact value of the integral. The Riemann sum was used in our exercise to approximate \( \sum_{k=1}^{n} \ln(1 + \frac{k}{n}) \), which can be represented as the integral \( \int_0^1 \ln(1+x) dx \). This plays a significant role in transitioning from a discrete sum to the corresponding integral, facilitating the calculation of the limit.

Logarithms in Calculus

Logarithms are incredibly powerful in calculus, especially for transforming products into sums. This transformation simplifies the problem of dealing with multiplicative sequences. When you take the logarithm of a product, such as \( \prod_{k=1}^{n} (1 + \frac{k}{n}) \), it allows you to express it as a sum: \( \sum_{k=1}^{n} \ln(1 + \frac{k}{n}) \).

This simplification is critical for handling complex products in limits or integrals. Understanding logarithms in calculus is essential to change the form of a difficult problem into something more straightforward to solve. By doing so, you can approach the problem step by step with easier calculations, leading to a clearer solution.

Integration by Parts

Integration by parts is a method derived from the product rule for differentiation. It is instrumental in finding the integral of products of functions. The formula is given by:\[\int u \ dv = uv - \int v \ du\]
In our exercise, integration by parts was employed to solve \( \int_0^1 \ln(1+x) dx \). Here, the selection of \( u \) as \( \ln(1+x) \) and \( dv \) as \( dx \) was strategic. It yielded simpler components to integrate, helping to solve what might seem like a complex integral initially.

The application of this technique underscores its value in calculus as it breaks down challenging integrals into manageable parts you can tackle bit by bit.

Exponential Function

The exponential function \( e^x \) holds a special place within calculus due to its unique properties, such as being equal to its own derivative. This property makes calculus operations involving \( e^x \) particularly seamless.

In solving the original limit problem, the exponential function \( \exp \left( \int_0^1 \ln(1+x) \ dx \right) \) was utilized to convert the result of the integral back into the product form raised to the power \( \frac{1}{n} \). This step is vital for determining the limit as part of a sequence or series.

The exponential function's relationship with logarithms is also significant because it allows for easy manipulation between sum and product forms, which is essential in these calculations.

Infinite Product

An infinite product is a product of an infinite number of factors. In calculus, infinite products often appear in the form of limits, sequences, or series. Computing their values involves understanding convergence - determining whether the sequence of partial products approaches a finite number.

Our exercise dealt with the limit of an infinite product \( \lim_{n \to \infty} \left[ \prod_{k=1}^{n} \left(1+\frac{k}{n}\right) \right]^{1/n} \). By taking logarithms, we converted this product into a limit of a sum, which could then be approximated by an integral. This technique allowed us to handle the infinite product effectively by transforming and simplifying it.

The concept of infinite products dovetails with the other ideas, such as exponential functions and logarithms, creating a toolkit for evaluating complex limits.