Problem 47 - A Complete Resource Book in Mathematics for JEE Main 2017

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 integer \( n \) is 3.

Step 1: Understand the Limits and Series Expansions

To solve this problem, we need to find the limit as \( x \rightarrow 0 \) of the given function, which involves \( \cos x \) and \( e^x \). It's useful to use the Taylor series expansions for \( \cos x \) and \( e^x \) while considering terms up to \( x^3 \) to determine the finite non-zero value of the limit.- The Taylor series expansion of \( \cos x \) up to \( x^3 \) is:\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \]- The Taylor series expansion of \( e^x \) up to \( x^3 \) is:\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]

Step 2: Substitute expressions into the limit

Substitute the series expansions into the original function:\[ (\cos x - 1)(\cos x - e^x) = \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - 1\right) \left(1 - \frac{x^2}{2} - \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right)\right) \]Simplifying within each parenthesis continues:- \( \cos x - 1 \approx -\frac{x^2}{2} + \frac{x^4}{24} \)- \( \cos x - e^x \approx -x - x^2 \) (considering terms up to \( x^3 \))

Step 3: Expand and simplify the product

Now, multiply these simplified expressions:\[(-\frac{x^2}{2} + \frac{x^4}{24})(-x - x^2) = \left(-\frac{x^2}{2}\right)(-x) + \left(-\frac{x^2}{2}\right)(-x^2) + \frac{x^4}{24}(-x) + \frac{x^4}{24}(-x^2)\]This simplifies to:\[\frac{x^3}{2} + \frac{x^4}{2} - \frac{x^5}{24} - \frac{x^6}{24}\]

Step 4: Identify the lowest power of x for finite limit

The expression simplifies primarily to \( \frac{x^3}{2} \) when adding terms for threshold around smallest contributing x power for non-zero finite limit: \[\frac{x^3}{2} = Cx^3 + O(x^4)\]The lowest power of \( x \) is \( x^3 \). Thus, to find the value of \( n \) such that \( \frac{f(x)}{x^n} \) results in a finite non-zero number, \( n \) must equal 3.

Key Concepts

Taylor Series

One of the most powerful tools in calculus is the Taylor series. It allows us to express functions as infinite sums of their derivatives at a single point. For functions like cosine and exponential functions, this can simplify complex problems, like the one given, into manageable algebra. The general form of a Taylor series expansion of a function \( f(x) \) centered at \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)

For our problem, we are especially interested in Taylor series expansions around \( x = 0 \), known as Maclaurin series. These involve substituting successive derivatives of the function at zero and are very useful for estimating function behavior around this point.
In the exercise solution, Taylor series help express \( \cos x \) and \( e^x \) in terms of powers of \( x \). In doing so, we can approximate these functions for very small \( x \), enabling us to perform precise calculations on their limits.

Cosine Function

The cosine function, symbolized as \( \cos x \), is one of the fundamental trigonometric functions. It relates the angle of a right triangle to the ratio of the adjacent side to the hypotenuse. In calculus, the behavior of \( \cos x \) as \( x \) approaches a specific value, such as zero, becomes critically important in limit calculations.
The Maclaurin series for \( \cos x \) is used to approximate its value near \( x = 0 \). For our exercise, the expansion is:

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

These terms allow us to substitute \( \cos x \) in our limit problem step by step, showing that when \( x \) is very small, \( \cos x \) is close to 1, with deviations captured by powers of \( x \). This understanding helps in simplifying expressions and directly affects how the problem resolves towards a finite limit.

Exponential Function

Exponential functions, particularly \( e^x \), hold special importance in mathematics. Known for their unique rate of growth, they appear in a variety of scenarios including calculus limit problems like this one. The Maclaurin series expansion for the exponential function is defined as:

• \( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \)

For the exercise we're solving, this series represents \( e^x \) as a sum of terms increasing with powers of \( x \). This form provides a way to understand how small changes in \( x \) affect the function’s value near zero.
By comparing \( e^x \) to \( \cos x \) and calculating expressions like \( \cos x - e^x \), we can zero in on simplified terms that are most significant. Understanding this expansion is crucial for applying limits and solving for a non-zero finite number.

Limits and Continuity

In calculus, analyzing the limit of a function as it approaches a specific value is central to understanding its behavior, especially near points where it could become undefined. Limits offer insights into function continuity and help find precise values and behaviors for otherwise complex expressions.
The exercise focuses on finding the limit of a rational expression involving \( \cos x \) and \( e^x \) over a power of \( x \) as \( x \) approaches zero. This scenario tests our ability to simplify and find which terms dominate the outcome as \( x \) becomes very small.

• Continuity ensures that small changes in \( x \) result in small changes in \( f(x) \), essential for finding precise limits.
• By simplifying using Taylor series and reducing terms, we identify the correct power \( n \) to yield a finite non-zero limit.

Understanding continuity and limits allows us to dissect problems into parts, enabling precise solutions like in this exercise where \( n \) equals 3 for a finite non-zero outcome.