Problem 7 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \(\underline{2}\) (B) \(\frac{e}{2}\) (C) \(\frac{e}{4}\) (D) \(\frac{4}{e}\)

Step-by-Step Solution

Answer

The answer is (A) \( \underline{2} \).

Step 1: Understand the Limit Expression

We are asked to evaluate \( \lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{1 / n} \). This represents the limit of a sequence of expressions raised to the power \( \frac{1}{n} \). Each term inside the bracket is of the form \( 1 + \frac{k}{n} \) where \( k \) takes values from 1 to \( n \). It essentially creates a product of \( n \) terms, each slightly above one.

Step 2: Rewrite the Product as an Exponential Form

Notice the expression: \( \left(1+\frac{k}{n}\right) = e^{\ln(1+\frac{k}{n})} \). The whole product can be represented as:\[ \left(1 + \frac{1}{n}\right) \left(1 + \frac{2}{n}\right) \cdots \left(1 + \frac{n}{n}\right) = e^{\sum_{k=1}^{n} \ln\left(1+ \frac{k}{n}\right)} \]

Step 3: Use the Approximation for Logarithms

For small \( x \), \( \ln(1 + x) \approx x - \frac{x^2}{2} + \cdots \). Since \( \frac{k}{n} \) is small as \( n \rightarrow \infty \), we approximate: \[ \ln(1 + \frac{k}{n}) \approx \frac{k}{n} \]This simplifies our expression. So the sum over logs can be approximated as:\[ \sum_{k=1}^{n} \frac{k}{n} = \frac{1}{n}\sum_{k=1}^{n} k = \frac{1}{2}(n+1) \approx \frac{n}{2} \]

Step 4: Insert the Approximated Sum Back into the Exponential Product

Recall that the product is \( e^{\sum_{k=1}^{n} \ln(1+\frac{k}{n})} \). With our approximation:\[ e^{\frac{n}{2}} \]

Step 5: Evaluate the Limit of the Resulting Expression

Now, we compute:\[ \lim_{n \to \infty} \left(e^{\frac{n}{2}}\right)^{1/n} = e^{\frac{1}{2}} \]

Step 6: Choose the Correct Answer

Comparing with provided options, \( e^{\frac{1}{2}} \) is equivalent to \( \sqrt{e} \), which matches the option \( \underline{2} \). Therefore, the correct answer is \( \underline{2} \).

Key Concepts

Product Notation

Product notation, represented by the capital Greek letter Pi (\(\prod\)), is a way to denote the repeated multiplication of a sequence of terms. This is similar to how summation notation (Sigma, \(\sum\)) is used for addition.
For example, the product \(\prod_{k=1}^{n} \left(1 + \frac{k}{n}\right)\) tells us to calculate the product of all terms of the form \(1 + \frac{k}{n}\) as \(k\) ranges from 1 to \(n\). This method condenses potentially long mathematical expressions into a more manageable form.

• Product notation is useful in sequences and series when evaluating complex expressions.
• It helps in spotting patterns and simplifying computations, especially in limits and asymptotic analysis.

In our exercise, the product notation signifies the multiplying of multiple terms, each slightly greater than one. This results in our expression capturing a cumulative growth effect, crucial for understanding sequences that extend to infinity.

Logarithmic Approximation

Logarithmic approximation is a technique used to simplify expressions involving logarithms, particularly when the input is close to one. The natural logarithm has a special property that allows us to approximate \(\ln(1+x)\) when \(x\) is very small.
Using the expansion \(\ln(1 + x) \approx x - \frac{x^2}{2} + \cdots\), when \(x\) is small, higher-order terms become negligible, and so \(\ln(1 + x) \approx x\).

• This approximation is powerful in simplifying calculations in limits and series.
• It transforms multiplicative processes into additive ones, which are often easier to handle.

In the context of the exercise, using \(\ln(1 + \frac{k}{n}) \approx \frac{k}{n}\) allows us to turn a complex exponential product into a much simpler summation. This approach is pivotal in finding the limit as \(n\) approaches infinity.

Exponential Functions

Exponential functions are crucial in mathematics due to their unique growth behaviors and properties, especially as their input values become very large or very small. The exponential function \(e^x\) grows rapidly as \(x\) increases and is uniquely defined by its own derivative being the function itself.
This makes exponential functions invaluable for continuous compounding in finance, natural processes like population growth, and, significantly, in analyzing limits and sequences.

• Exponential functions turn addition in the exponent into multiplication outside of it, which is key for simplifying products of many terms.
• They maintain positivity and never reach zero, which is crucial for many real-world applications.

In our step-by-step solution, expressing the product as \(e^{\sum_{k=1}^{n} \ln(1+\frac{k}{n})}\) showcases this power. By transforming a product into an exponential function, we simplify the process of taking the limit of sequences.

Asymptotic Analysis

Asymptotic analysis is a method used to describe the behavior of functions as inputs become very large. It involves comparing functions and understanding their limits, which helps predict the general trend without evaluating every detail.
It is often applied in algorithms, physics, and many areas where exact calculations can be impractical. Asymptotic analysis simplifies understanding of limits by approximating expressions, offering insights into their eventual outcome as inputs head towards infinity.

• Focusing on dominant terms in functions helps with practical computations.
• It aids in comprehending the growth rate and convergence properties of sequences and series.

In the exercise, employing asymptotic approximations helps describe how the logarithmic sum \(\frac{n}{2}\) behaves as \(n\) becomes large, ultimately simplifying the limit evaluation process. This method is essential in efficiently solving complex problems that involve infinity.