Problem 24 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{1-2+3-4+5-6+\ldots-2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}\) is equal to (A) \(\frac{1}{3}\) (B) \(-\frac{1}{3}\) (C) \(-\frac{1}{5}\) (D) None of these

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{3}\), so the answer is (B).

Step 1: Simplify the Series

Given the series in the numerator: \(1-2+3-4+5-6+\ldots-2n\), observe that it can be grouped as \((1-2) + (3-4) + \ldots + ((2n-1) - 2n)\). Each pair \((k) - (k+1)\) equals -1. Thus, with pairs until \(2n\), there are \(n\) terms. Therefore, the sum in the numerator is \(-n\).

Step 2: Simplify the Denominator

Consider the denominator \(\sqrt{n^{2}+1} + \sqrt{4n^{2}-1}\). Use the approximation \(\sqrt{a^2+b}\approx a + \frac{b}{2a}\) for large \(a\). Here, \(\sqrt{n^2 + 1} \approx n + \frac{1}{2n}\) and \(\sqrt{4n^2 - 1} \approx 2n - \frac{1}{4n}\). Therefore, the denominator is approximately \(n + \frac{1}{2n} + 2n - \frac{1}{4n} = 3n\).

Step 3: Calculate the Limit

Now, evaluate the limit: \[\lim_{n \to \infty} \frac{-n}{3n}\]This simplifies to \(-\frac{1}{3}\) as all terms of \(n\) cancel out except for the constants.

Step 4: Identify the Correct Choice

The limit evaluates to \(-\frac{1}{3}\). Check the given options. The correct option is (B) \(-\frac{1}{3}\).

Key Concepts

Infinite Series

An infinite series is the sum of terms in an infinite sequence. In our problem, the series in the numerator is a combination of numbers alternating between positive and negative signs. This series takes the form:

• 1 - 2 + 3 - 4 + 5 - 6 + ...

By grouping the terms as

• (1 - 2) + (3 - 4) + (5 - 6)...

each pair equals -1. Since the series continues up to 2n, this means there are n pairs in total. Thus, the sum becomes

• -n.

Understanding how to manipulate such series is crucial when working with limits and infinite sequences. This manipulation helps simplify expressions, facilitating the evaluation of their limits.

Square Root Approximation

When evaluating limits involving square roots, approximations can greatly simplify the calculation. For large numbers, approximations help you see the behavior as the variable approaches infinity. The general approximation for a square root is:

• \( \sqrt{a^2 + b} \approx a + \frac{b}{2a} \)

In our problem, applying this approximation method simplifies the expression inside the square root.

• For \( \sqrt{n^2 + 1} \), approximate as \( n + \frac{1}{2n} \)
• For \( \sqrt{4n^2 - 1} \), approximate as \( 2n - \frac{1}{4n} \)

Adding these gives us approximately 3n, making it easier to evaluate the limit. Understanding square root approximation is an essential tool, especially for limits with large variable expressions.

Limit Evaluation

Limits are a fundamental tool in calculus for understanding the behavior of functions as they approach a particular point or infinity. To evaluate the limit in our example:

• \[ \lim_{n \to \infty} \frac{-n}{3n} \]

We simplify by canceling out the terms, focusing on the coefficients. Since both the numerator and denominator share \( n \), it simplifies directly to:

• \(-\frac{1}{3} \)

Limit evaluation is about manipulation and simplification. It's crucial to identify cancelling opportunities to isolate terms that truly impact the value of the limit. Once simplified, checking these against multiple-choice options solidifies understanding and application of limits in solving calculus problems.