Problem 47 - A Complete Resource Book in Mathematics for JEE Main

Question

The integer \(n\) for which \(\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^{x}\right)}{x^{n}}\) is a finite non-zero number, is (A) 1 (B) 2 (C) 3 (D) 4

Step-by-Step Solution

Answer

The correct answer is (C) 3.

Step 1: Identify the Expression Inside the Limit

The given limit is \( \lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^{x})}{x^{n}} \). We need to find the smallest integer \( n \) that makes this limit finite and non-zero as \( x \rightarrow 0 \).

Step 2: Use Taylor Series Expansions

We expand the functions \( \cos x \) and \( e^{x} \) using their Taylor series around 0.\[ \cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \]\[ e^{x} \approx 1 + x + \frac{x^2}{2} + \cdots \].

Step 3: Simplify Using Leading Terms

Substitute the Taylor series into the expression:\[ \cos x - 1 \approx -\frac{x^2}{2}, \quad \cos x - e^{x} \approx - \left( x + \frac{x^2}{2} \right).\]Multiplying these gives:\[ (\cos x - 1)(\cos x - e^{x}) \approx \left(-\frac{x^2}{2}\right)\left(-x - \frac{x^2}{2}\right) = \frac{x^3}{2} + \frac{x^4}{4}. \]

Step 4: Determine the Leading Order Term

The leading term in the product expansion is \( \frac{x^3}{2} \). Therefore, the expression in the numerator behaves like \( \frac{x^3}{2} \) for \( x \to 0 \).

Step 5: Find the Value of n

To make the limit \( \lim _{x \rightarrow 0} \frac{x^3/2}{x^{n}} \) finite and non-zero, the exponent of \( x \) in the numerator and denominator must be equal. Thus \( n \) must be 3 since the leading term in the numerator is \( x^3 \).

Step 6: Conclusion

Hence, \( n = 3 \) is the smallest integer such that the limit is finite and non-zero.

Key Concepts

Taylor Series Expansion

The Taylor series is a powerful tool in calculus, used to approximate functions by expanding them into an infinite sum of terms calculated from the function's derivatives at a single point. In the exercise, Taylor series are used to approximate the functions \( \cos x \) and \( e^x \) around \( x = 0 \), which is also referred to as a Maclaurin series. This approximation allows us to express complex functions in a simpler polynomial form.

For \( \cos x \), the Taylor series expansion is:

• \( \cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \)

For \( e^x \), the series is:

• \( e^x \approx 1 + x + \frac{x^2}{2} + \cdots \)

By substituting these expansions into the expression \((\cos x - 1)(\cos x - e^x)\), we simplify the function using only the leading terms. This process highlights the primary contributors to the function's behavior as \( x \) approaches zero.

The advantage of using Taylor series here is that it reduces the complexity of trigonometric and exponential functions, making it easier to analyze the limit behavior as \( x \to 0 \).

Trigonometric Functions

Trigonometric functions like \( \cos x \) are periodic functions that play a key role in mathematics, especially in calculus and analysis. They relate angles of triangles to the lengths of their sides and are extensively used to model periodic phenomena such as sound and light waves.

In this exercise, we focus on the Taylor series expansion of \( \cos x \) around \( x = 0 \), expressed as:

• \( \cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \)

This series helps us analyze small-angle behaviors, crucial for evaluating limits as \( x \) approaches zero.

When \( \cos x \) is near zero angles, the series simplifies significantly, allowing the approximation \( \cos x \approx 1 - \frac{x^2}{2} \), which is incredibly useful in limit problems. The higher-order terms \( \frac{x^4}{24} + \cdots \) become insignificant for \( x \) close to zero, simplifying the calculation of limits. This simplification transforms a usually complex trigonometric function into something much more manageable and intuitive.

Exponential Functions

Exponential functions, such as \( e^x \), are characterized by their constant rate of growth, making them fundamental in both mathematics and natural sciences. They describe many natural phenomena and processes, like population growth and radioactive decay.

In the context of limits and functions approaching a point, understanding the behavior of \( e^x \) near zero is crucial. Its Taylor series expansion at \( x = 0 \) is written as:

• \( e^x \approx 1 + x + \frac{x^2}{2} + \cdots \)

This expression illustrates how \( e^x \) behaves around small \( x \), where the function can be approximated closely by few terms.

This exercise shows that when \( x \to 0 \), you can simplify calculations by using just the primary terms of the Taylor series. This simplification makes it easier to find the leading term in our expression for limit evaluation. By disregarding the higher powers of \( x \), such as \( x^2/2 \) and beyond, we maintain accuracy while simplifying the algebra involved in processing limits.