Problem 164 - A Complete Resource Book in Mathematics for JEE Main

Question

We know that for a continuous function \(f\) in \([a, b]\) $$ \lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)=\int_{a}^{b} f(x) d x, h=\frac{b-a}{n} $$ On putting \(a=0, b=1 ;\) (1) becomes $$ \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right)=\int_{0}^{1} f(x) d x $$ This formula enables us to evaluate limits of the form: $$ \lim _{n \rightarrow \infty}\left[\phi_{1}(n)+\phi_{2}(n)+\ldots+\phi_{n}(n)\right] $$ To evaluate this limit we express the \(r\) th term as $$ \phi_{r}(n)=\frac{1}{n} f\left(\frac{r}{n}\right) $$ and then replace \(\frac{r}{n}\) by \(x, \frac{1}{n}\) by \(d x\) and \(\lim _{n \rightarrow \infty} \sum_{r=1}^{n}\) by \(\int_{0}^{1}\) Also, \(\lim _{n \rightarrow \infty} \sum_{r=1}^{k n} \frac{1}{n} f\left(\frac{r}{n}\right)=\int_{0}^{k} f(x) d x\) \(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{1 / n}\) is equal to (A) \(\frac{2}{-}\) (B) \(\frac{e}{2}\) (C) \(\frac{e}{4}\) (D) \(\frac{4}{e}\)

Step-by-Step Solution

Answer

(B) \(\frac{e}{2}\)

Step 1: Recognize the Structure of the Limit

The problem asks for the limit of a product as \( n \to \infty \), suggesting we might express it in an exponential form suitable for manipulation via standard calculus tools. The mathematical expression is \( \lim_{n \to \infty} \left[ \phi_1(n) \phi_2(n) \cdots \phi_n(n) \right]^{1/n} \), which involves a repeated product.

Step 2: Express the Product in Logarithmic Form

To evaluate the limit of a product, take the natural logarithm to transform it into a sum: \( \ln L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left(1 + \frac{r}{n}\right) \). This makes it easier to handle, since sums are generally more analysable than products.

Step 3: Use Limit Definition to Generate Riemann Sum

The expression \( \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left(1 + \frac{r}{n}\right) \) fits the structure of a Riemann sum for the integral \( \int_{0}^{1} \ln(1 + x) \, dx \) when \( \frac{r}{n} \to x \) and \( \frac{1}{n} \to dx \). Here, \( f\left(\frac{r}{n}\right) = \ln \left(1 + \frac{r}{n}\right) \).

Step 4: Compute the Integral

To find the value of \( \int_{0}^{1} \ln(1 + x) \, dx \), perform integration using integration by parts or a substitution if needed. Doing integration by parts with \( u = \ln(1+x) \) and \( dv = dx \), we obtain \( \int_{0}^{1} \ln(1 + x) \, dx = \frac{3}{4} - \ln(2) \).

Step 5: Interpret the Final Result

Therefore, \( \ln L = \frac{3}{4} - \ln(2) \), and to find \( L \), take the exponential: \( L = e^{\frac{3}{4} - \ln(2)} = \frac{e^{3/4}}{e^{\ln(2)}} = \frac{e^{3/4}}{2} \). Hence, \( L = \frac{e^{3/4}}{2} \).

Key Concepts

Continuous Function

In mathematics, a continuous function plays a vital role in understanding the behavior of real-valued functions. Simply put, a function is considered continuous at a point if there are no sudden jumps or breaks in its value. Imagine tracing the line or curve of the function on a graph without lifting your pencil—this smooth journey signifies continuity.

A function is typically continuous over an interval if it is continuous at each point within that interval. Consider a function like \(f(x)=\sin(x)\) over the interval \( [a, b] \) — it's continuous because \(\sin(x)\) varies without breaks from \(-1\) to \(1\).

In the realm of Riemann sums and integration, continuity ensures the existence of a definite integral over an interval. Without it, calculating integrals becomes complex, as discontinuities prevent the smooth summing of "infinitely small" parts of the function's behavior. Hence, when computing limits and integrals, it's essential to first confirm that the function involved is continuous over the relevant interval.

Limits

The concept of limits forms the backbone of calculus, acting as the foundation for defining derivatives and integrals. A limit helps us understand how a function behaves as it approaches a particular point or value. It's like inching closer to a barrier—while never quite reaching it but understanding what lies beyond.

In mathematical notation, \(\lim_{n \to \infty}\) defines the behavior of a function as the parameter \(n\) grows infinitely large. Using Riemann sums, we see this concept in practice as the expression approaches an integral. For example, the Riemann sum \(\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right)\)transitions into \(\int_a^b f(x) \, dx\), illustrating how discrete summation approximates continuous integration when \(n\) becomes very large.

Understanding limits helps us deal with functions that exhibit challenging properties as they stretch towards infinity or shrink towards zero, giving meaning and precision to the concept of infinitely small or large quantities.

Integration by Parts

Integration by parts is a powerful technique in calculus used to integrate products of functions. It is derived from the product rule of differentiation and offers an efficient strategy to break down complex integrals, particularly when one part is easily integrable compared to its counterpart.

The formula for integration by parts is derived from the product rule: \(\int u \, dv = uv - \int v \, du\). Here, you choose one function as \(u\) to differentiate, and another as \(dv\) to integrate. This choice is crucial—generally, selecting \(u\) as a function that becomes simpler upon differentiation improves the process.

For example, in the integral \(\int_{0}^{1} \ln(1+x) \, dx\), using integration by parts requires setting \(u = \ln(1+x)\) and \(dv = dx\). Upon differentiating and integrating, we simplify and compute the definite integral, resulting in \(\frac{3}{4} - \ln(2)\). This showcases how integration by parts transforms a daunting problem into manageable steps, broadening the scope of what we can compute analytically.