Problem 5 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \cos \left(\pi \sqrt{n^{2}+n}\right), n \in Z\) is equal to (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

0

Step 1: Simplifying the Square Root Expression

First, we recognize that \( n^2 + n \) is inside the square root. We can approximate \( \sqrt{n^2 + n} \) for large \( n \) as follows:\[\sqrt{n^2 + n} = \sqrt{n^2(1 + \frac{1}{n})} = n\sqrt{1 + \frac{1}{n}}.\] For \( n \to \infty \), \( \sqrt{1 + \frac{1}{n}} \approx 1 + \frac{1}{2n} \) using the binomial expansion approximation.

Step 2: Approximating the Expression Inside Cosine

By using the approximation \( \sqrt{1 + \frac{1}{n}} \approx 1 + \frac{1}{2n} \), we expand the expression inside cosine as:\[\pi \sqrt{n^2 + n} = \pi n \sqrt{1 + \frac{1}{n}} \approx \pi n \left(1 + \frac{1}{2n}\right) = \pi n + \frac{\pi}{2}.\]

Step 3: Evaluating the Limit

Now we need to find \( \lim_{n \to \infty} \cos\left(\pi n + \frac{\pi}{2}\right) \).We know that\[\cos(\pi n + \frac{\pi}{2}) = \cos\left(\pi n\right)\cos\left(\frac{\pi}{2}\right) - \sin(\pi n)\sin\left(\frac{\pi}{2}\right).\] Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), it simplifies to:\[-\sin(\pi n).\]

Step 4: Analyzing \( \sin(\pi n) \) for Integer \( n \)

Considering \( n \in Z \), \( \sin(\pi n) = \sin k\pi \) where \( k \) is an integer. Here, \( \sin(k\pi) = 0 \) always, for any integer \( k \).Therefore, \(-\sin(\pi n) = -0 = 0\).

Key Concepts

Trigonometric Limits

Trigonometric limits are a fundamental part of calculus as they help determine how functions that involve trigonometric components, like sine and cosine, behave as they approach a particular point or infinity. When dealing with trigonometric limits, key trigonometric identities and properties become essential. For instance, one of the essential trigonometric identities is

• the Pythagorean identity: \( \sin^2(x) + \cos^2(x) = 1 \).
• Another frequently used principle is the behavior of trigonometric functions as their arguments increase or repeat cyclically. The sine and cosine functions are periodic with period \( 2\pi \), meaning that they repeat every full rotation of a circle — every \( 2\pi \) radians.

This means when evaluating limits such as \( \lim_{n \to \infty} \cos(\pi n + C) \), where \( C \) is a constant, it’s often about determining the function's periodic result.Trigonometric identities help simplify complex expressions — for example, in this exercise, leveraging \( \cos(\frac{\pi}{2}) = 0 \) led us to focus on \(-\sin(\pi n)\), which consistently evaluates to zero for integer values, since \( \sin(k\pi) = 0 \) for any integer \( k \). This makes understanding trigonometric limits a blend of identities and behavior comprehension.

Infinite Limits

Infinite limits look at the behavior of a function as the variable grows arbitrarily large or decreases without bound. In simpler terms, we are interested in what value a function approaches as the variable tends towards infinity. When we write \( \lim_{n \rightarrow \infty} \) or \( \lim_{x \rightarrow -\infty} \), it signals that we’re checking the trend or asymptote in function behavior rather than its exact value at any particular point.In our specific example, we analyzed the limit of a trigonometric function involving a square root — a classic scenario where infinite limits arise. As \( n \to \infty \), we first isolated the square root \( \sqrt{n^2 + n} \) and approximated it using notable mathematical techniques such as the binomial expansion to simplify and understand its behavior better:

• \( \sqrt{n^2 + n} \approx n\left(1 + \frac{1}{2n}\right) \).

Such approximations simplify the algebra tied within. Thus, determining what \( n\sqrt{1 + \frac{1}{n}} \) approaches as \( n \to \infty \) aids in predicting the overall trend of the expression inside the trigonometric function. Understanding infinite limits equips you with the foresight into how functions behave at a large scale, especially regarding tendencies rather than concrete values alone.

Asymptotic Analysis

Asymptotic analysis is a powerful tool in calculus that helps us understand how functions behave as the input grows very large. This type of analysis allows mathematicians to make approximations that simplify complex expressions. Typically involving identifying dominant terms that ultimately influence how functions typically act as variables increase towards infinity.A typical approach is to re-write complicated expressions in simpler forms, focusing only on terms with significant impact as the variables grow. For example, in the problem at hand, we simplified \( \sqrt{n^2 + n} \) into a more manageable form using the approximation:

• \( n\sqrt{1 + \frac{1}{n}} = n \left(1 + \frac{1}{2n} \right) \).

Dropping insignificant terms—like those that diminish in comparison to others—as \( n \) gets very large is the key practice in asymptotic analysis. Such simplifications help in determining the eventual behavior of more complex mathematical entities.Furthermore, understanding periodicity and behavior of functions such as \( \sin \) and \( \cos \), specifically when dealing with added constants like \( \pi/2 \), allows for judging how their cycle and behavior converge. Asymptotic analysis does not just simplify; it provides a way to analyze limits, behaviors, and eventual tendencies effectively.