Problem 72 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 0} \frac{\sqrt{1-\sqrt{\cos x}}}{x}=\) (A) \(\frac{1}{2}\) (B) \(-\frac{1}{2}\) (C) Does not exist (D) None of these

Step-by-Step Solution

Answer

(A) \(\frac{1}{2}\)

Step 1: Understand the Problem

We need to find the limit of \( \frac{\sqrt{1-\sqrt{\cos x}}}{x} \) as \( x \to 0 \). This involves evaluating the limit of a complex expression which includes a square root and a cosine function.

Step 2: Use Taylor Expansion for \( \cos x \)

Apply the Taylor series expansion for \( \cos x \) near 0. We have \( \cos x \approx 1 - \frac{x^2}{2} \) for small \( x \). Substituting in the expression gives \( \sqrt{1-\sqrt{1 - \frac{x^2}{2}}} \).

Step 3: Simplify the Inner Square Root

For small \( x \), approximate \( \sqrt{1 - \frac{x^2}{2}} \) using the expansion \( \sqrt{1+y} \approx 1 + \frac{y}{2} \) for small \( y \). Thus, \( \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{4} \).

Step 4: Simplify the Entire Expression

Substitute the approximation back: \( 1 - (1 - \frac{x^2}{4}) \approx \frac{x^2}{4} \). So \( \sqrt{1-\sqrt{1 - \frac{x^2}{2}}} \approx \sqrt{\frac{x^2}{4}} = \frac{x}{2} \).

Step 5: Evaluate the Limit

Substitute \( \frac{x}{2} \) back into the original limit to get \( \lim_{x \to 0} \frac{\frac{x}{2}}{x} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \).

Step 6: Conclusion

Thus, the limit is \( \frac{1}{2} \). Therefore, option (A) is correct.

Key Concepts

Taylor series expansion

When faced with complicated functions in calculus, one useful tool is the Taylor series expansion. This allows us to approximate functions with polynomials, making them easier to manipulate and integrate.

For instance, the function \(cos x\) can be approximated around zero as \(cos x \approx 1 - \frac{x^2}{2}\).This approximation is particularly useful when \(x\) is small, helping simplify trigonometric expressions into polynomial forms.

• Taylor series can convert complex functions to polynomials.
• Especially beneficial for functions near zero or other points.
• Provides a method to find limits and approximations.

Using Taylor expansions gives us a way to simplify the problem by making functions like square roots and trigonometric functions more manageable, allowing us to focus on simpler polynomial mathematics.

Trigonometrical functions

Trigonometric functions, such as cosine, are fundamental in calculus, especially when dealing with periodic patterns and waveforms. Understanding these functions and their properties is crucial for evaluating limits and other calculus operations.

In the given problem, we use the cosine function, \(\cos x\), which oscillates between -1 and 1 and has known properties around small angles. When \(x\) is near zero:

• \(\cos x \approx 1 - \frac{x^2}{2}\): Used for easier calculations.
• This approximation simplifies complex expressions, like square roots of \(\cos x\).

This form simplifies the cosine in expression, letting us focus on small polynomials in limit problems. This simplification is key to many solutions in calculus involving limits.

Square root functions

Square root functions often appear in calculus problems, especially when dealing with limits and continuous functions.

In such cases, especially when under a square root or nested functions, they can complicate limits.

• We can use approximations like \(\sqrt{1 + y} \approx 1 + \frac{y}{2}\) for small \(y\).
• Simplifies expressions to aid in evaluating limits.

In the problem, we substitute \(\sqrt{1 - \cos x}\) with its polynomial approximation. It becomes approachable when the inner functions are simplified into polynomials, thus enabling us to evaluate complex nested functions efficiently. Understanding how and when to use these approximations is vital in tackling calculus problems involving square root operations.