Problem 43 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{x \rightarrow \infty} \sqrt{\frac{x+\sin x}{x-\cos x}}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The limit is 1, so the answer is (B).

Step 1: Identify the Limit Expression

Recognize that we need to find the limit of the expression \( \lim_{x \to \infty} \sqrt{\frac{x+\sin x}{x-\cos x}} \). This involves analyzing the behavior of the function as \( x \) approaches infinity.

Step 2: Simplify the Expression

To simplify, observe that as \( x \to \infty \), \( \sin x \) and \( \cos x \) are bounded between -1 and 1, and are negligible compared to \( x \). Thus, approximate \( x + \sin x \approx x \) and \( x - \cos x \approx x \).

Step 3: Write the Simplified Limit

Use the approximations to simplify the expression inside the square root: \( \frac{x + \sin x}{x - \cos x} \approx \frac{x}{x} = 1 \). The limit now becomes \( \lim_{x \to \infty} \sqrt{1} \).

Step 4: Calculate the Limit

Since \( \sqrt{1} = 1 \), the limit is \( \lim_{x \to \infty} 1 = 1 \).

Step 5: Conclusion

Therefore, based on our calculations and simplifications, the limit is \( \boxed{1} \). The correct answer is option (B) 1.

Key Concepts

Infinite Limits

Infinite limits are an important concept in calculus where we study the behavior of functions as they approach infinity. In these scenarios, the variable approaches an infinite value, and the goal is to determine whether the function approaches a specific number or becomes unbounded. This is crucial for understanding the long-term behavior of functions.

When faced with limits heading towards infinity, the challenge often lies in simplifying complex expressions so they can be more easily evaluated.
In our exercise, as \( x \) tends to infinity, the terms \( \sin x \) and \( \cos x \), which oscillate between \(-1\) and \(1\), become negligible in comparison to \( x \).
This simplification allows us to focus on the most dominant term, which is \( x \) in both the numerator and the denominator of the given expression, making it easier to find the limit.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are periodic with outputs ranging between -1 and 1. This periodicity and bounded nature make these functions particularly interesting when evaluating limits.

Although they undergo continuous changes, their oscillating behavior is predictable, especially at large values of \( x \). In our context of taking limits, the effect of \( \sin x \) or \( \cos x \) becomes less significant relative to other changes in the function.
This is because as \( x \) approaches infinity, the behavior of the function \( \sin x \) and \( \cos x \) is overshadowed by \( x \), the linearly growing term.
This is why we consider \( x + \sin x \approx x \) and \( x - \cos x \approx x \), which simplifies our work significantly.

Limit Simplification

Limit simplification is a powerful technique where we reduce complex limit expressions to simpler forms that are more manageable to evaluate.

Simplifying limits often involves factoring out dominant terms and recognizing which terms can be considered negligible. This matches closely with algebraic simplification techniques.
In our exercise, simplifying the expression \( \frac{x + \sin x}{x - \cos x} \) to \( \frac{x}{x} \) is a crucial step.
By realizing that both \( x + \sin x \) and \( x - \cos x \) have a dominant \( x \), we can see that the other terms contribute very little as \( x \to \infty \).
This simplifies the expression to \( \frac{x}{x} = 1 \), leading directly to finding \( \lim_{x \to \infty} \sqrt{1} \). Simplifications help in uncovering the true behavior of functions under extreme conditions.