Problem 13 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(x_{n}=\frac{1-2+3-4+5-6+\ldots \ldots \ldots 2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}\), then \(\lim _{n \rightarrow \infty} x_{n}\) is equal to (a) \(1 / 3\) (b) \(-1 / 3\) (c) \(2 / 3\) (d) \(-2 / 3\)

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{3}\) (option b).

Step 1: Understand the Pattern in the Numerator

The numerator of the given expression is a sequence: 1, -2, 3, -4, ..., 2n. Notice that the pattern alternates adding odd integers and subtracting even integers. Therefore, the sum of the first 2n terms can be rewritten as \( S = 1 - 2 + 3 - 4 + \ldots + (2n-1) - 2n \). Observe that pairs like \((1 - 2), (3 - 4), \ldots, ((2n-1) - 2n)\) result in -1, repeated n times, hence \( S = -n \).

Step 2: Simplify the Denominator

The denominator is \( \sqrt{n^2 + 1} + \sqrt{4n^2 - 1} \). Approximating for large n, notice \( \sqrt{n^2 + 1} \approx \sqrt{n^2} = n \) and \( \sqrt{4n^2 - 1} \approx \sqrt{4n^2} = 2n \). So, the denominator approximates to \( n + 2n = 3n \).

Step 3: Calculate the Expression

Now that we have the simplified numerator and denominator, the expression becomes \( x_n = \frac{-n}{3n} = -\frac{1}{3} \).

Step 4: Evaluate the Limit

As \( n \to \infty \), the expression \( x_n = -\frac{1}{3} \) remains constant. Therefore, \( \lim_{n \to \infty} x_n = -\frac{1}{3} \).

Key Concepts

Understanding Sequence Patterns in the Numerator

When tackling problems like this, recognizing patterns in sequences is very helpful. In the given problem, you see the sequence as 1, -2, 3, -4, ..., 2n. This sequence alternates between adding odd numbers and subtracting even numbers. If you pair up consecutive terms such as (1 - 2), (3 - 4), you get pairs that always sum up to -1. By identifying this, you can see that for each complete pair from 1 to 2n, you are effectively subtracting 1, over and over. This means the entire sequence up to 2n equals \(-n\), because there are \(n\) such pairs. It's all about spotting these sequences that repeat themselves in a structured pattern.

• Pair up terms: \((1 - 2), (3 - 4), \,\ldots\, ((2n-1) - 2n)\)
• Each pair sums to -1
• Total sum for 2n terms: \(-n\)

Understanding Infinity in Calculus For Limits

With calculus, you often look at how expressions behave as they head towards infinity. Let's break this down using the denominator of our function which is \( \sqrt{n^2 + 1} + \sqrt{4n^2 - 1} \). When variables like \(n\) grow large, terms within square roots simplify significantly. A key trick is to focus on the term inside the root that's going to dominate when \(n\) is very large. For \( \sqrt{n^2 + 1}\), as \(n\) becomes huge, \(\sqrt{n^2 + 1} \approx n\). Similarly, \(\sqrt{4n^2 - 1} \approx 2n\). The point with these limits is that you can often ignore the smaller parts of an expression that become negligible.

• For large \(n\), focus on the term that's biggest.
• \(\sqrt{n^2 + 1} \to n\) and \(\sqrt{4n^2 - 1} \to 2n\).
• The limit helps us simplify and find behavior trends.

Numerator and Denominator Simplification

Simplifying complex expressions involves making things as simple as possible without changing the value. In this case, after understanding both the numerator and denominator, we plug them back into the expression \(x_n = \frac{-n}{3n}\). The numerator \(-n\) is derived from the sequence pattern, while the denominator \(3n\) comes from the simplification involving infinity in calculus.Once you substitute these, the expression \(x_n\) simplifies to \(-\frac{1}{3}\). This simplification shows how mathematical expressions become easier to understand using orderly techniques.

• Numerator from sequence: \(-n\)
• Denominator simplification: \(3n\)
• Expression simplifies to consistent value: \(x_n = -\frac{1}{3}\)

Subsequently, when taking the limit as \(n\) approaches infinity, the problem becomes much easier to evaluate. Continuous simplification ensures a constant value at infinity, which is crucial in calculus.