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

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

The limit does not exist as it oscillates between 1 and -1.

Step 1: Simplify the Expression Inside the Cosine

We start by simplifying the expression under the square root: \[ \pi \sqrt{n^2 + n} = \pi \sqrt{n^2(1 + \frac{1}{n})}. \]This can be rewritten as:\[ \pi \sqrt{n^2} \sqrt{1 + \frac{1}{n}} = \pi n \sqrt{1 + \frac{1}{n}}. \]

Step 2: Find the Limit of the Simplified Expression

As \( n \rightarrow \infty \), the term \( \frac{1}{n} \rightarrow 0 \). Therefore,\[ \sqrt{1 + \frac{1}{n}} \rightarrow \sqrt{1} = 1. \]Thus, \[ \pi n \sqrt{1 + \frac{1}{n}} \rightarrow \pi n. \]

Step 3: Analyze the Behavior of Cosine

Our expression becomes \( \cos(\pi n) \). The cosine function has the property that when \( n \) is an integer:- \( \cos(\pi n) = (-1)^n \).This is because \( \pi n \) is an odd multiple of \( \pi \) when \( n \) is odd and an even multiple when \( n \) is even.

Step 4: Determine the Limit

In even cases, \( (-1)^n = 1 \), and in odd cases, \( (-1)^n = -1 \). However, as \( n \to \infty \), the behavior of cosine does not settle in at any single value; it oscillates forever between 1 and -1.

Key Concepts

Asymptotic Behavior

Asymptotic behavior examines how sequences behave as they grow towards infinity. As sequences extend further, they tend to get closer to a predictable value, but in some cases, like oscillating sequences, no single value is reached. In our exercise, the limit of the sequence \( \lim _{n \rightarrow \infty} \cos(\pi \sqrt{n^2 + n}) \) showcases such an oscillating behavior. Here, the sequence alternates between 1 and -1 due to the properties of the cosine function applied to \( \pi n \).
Understanding asymptotic behavior helps us determine overall trends and stabilize sequence predictions.
It tells us that as \( n \to \infty \), our sequence's behavior doesn't fixate on any single number due to the ongoing alternation. This indicates that the limit is not a single real number, thus the correct answer to the exercise is (D) None of these.

Integer Sequences

When dealing with integer sequences, like \( n \in \mathbb{Z} \), each term is a whole number. In many mathematical problems, integers are used to denote terms that sequentially continue indefinitely as \( n \to \infty \).
The importance of recognizing integer sequences lies in correctly applying mathematical operations and properties. For example, in our exercise, knowing that \( n \) is an integer is crucial for evaluating \( \cos(\pi n) \) because the cosine values depend on whether \( n \) is even or odd.

• The sequence of integers progresses through natural numbers, \( 0, 1, 2,... \).
• Each increase in \( n \) causes a corresponding calculation in the sequence function. Although the computations may change drastically, integer index sequences are predictable and orderly enough to allow effective mathematical analysis.

Cosine Function Properties

The cosine function, \( \cos(\theta) \), is a periodic function with a key property of oscillation between -1 and 1. This oscillatory nature emerges prominently when the function input is based on multiples of \( \pi \), as seen in our exercise.
When multiplied by \( \pi \), the cosine function dictates:

• \( \cos(\pi n) = 1 \) when \( n \) is even.
• \( \cos(\pi n) = -1 \) when \( n \) is odd.

This alternation leads to an oscillating sequence as \( n \) increases, which does not converge to a single limit.
The properties of the cosine function help us in problems like these to identify the overall behavioral pattern anticipated. The function influences entire sequence behaviors, making cosine a vital component in understanding oscillatory dynamics in sequences. Thus, recognizing that alternating pattern is key to solving this mathematical sequence problem.