Problem 112 - A Complete Resource Book in Mathematics for JEE Main

Question

We know that if \(\lim _{x \rightarrow a} f(x)=l\) and \(\lim _{x \rightarrow a} g(x)=m(\neq 0)\), then $$ \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)} $$ However, if \(\lim _{x \rightarrow a} g(x)=0=\lim _{x \rightarrow a} f(x)\), we cannot say anything definite about the existence of \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\). Though in some cases this limit exists. Any expression of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) is termed as an indeterminate form. Many other expressions like \(\infty-\infty, 1^{\infty}, \infty^{0}, 0^{0}, 0 \times \infty\) which can be reduced to the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) are also called indeterminate forms. If \(\frac{f(x)}{g(x)}\) is indeterminate at \(x=a\) of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then $$ \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ where \(f^{\prime}\) is derivative of \(f\). If \(\frac{f^{\prime}(x)}{g^{\prime}(x)}\), too, is indeterminate at \(x=a\) of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim _{x \rightarrow a} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}\) This can be continued till we finally arrive at a determinate result. \(\lim _{x \rightarrow 0}|x|^{\sin x}\) equals (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

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

Step 1: Understand the Limit Expression

We need to evaluate the limit \( \lim_{x \rightarrow 0}|x|^{\sin x} \). This expression involves both an absolute value and an exponent that varies with \( x \).

Step 2: Transform the Expression for Simplification

Consider the expression in a logarithmic form for simplification: \( y = |x|^{\sin x} \). Taking natural logs on both sides gives \( \ln y = \sin x \cdot \ln |x| \). This transformation will help us handle the exponent more easily.

Step 3: Analyze the New Limit

We now need to find \( \lim_{x \rightarrow 0} \sin x \cdot \ln |x| \). This is a product of functions that become \(0 \cdot (-\infty)\) as \( x \rightarrow 0 \), which is an indeterminate form.

Step 4: Apply L'Hôpital's Rule to Resolve Indeterminacy

Rewrite \( \sin x \cdot \ln |x| \) as \( \frac{\ln |x|}{1/(\sin x)} \). Now, this limit is of the form \( \frac{\infty}{\infty} \), which allows us to use L'Hôpital's Rule.

Step 5: Differentiate and Simplify

Differentiate the numerator and denominator: \( \frac{d}{dx}(\ln |x|) = \frac{1}{x} \) (for \( x eq 0 \)), and \( \frac{d}{dx}\left( \frac{1}{\sin x} \right) = \frac{-\cos x}{\sin^2 x} \). Now, find \( \lim_{x \rightarrow 0} \frac{1/x}{-\cos x/\sin^2 x} \).

Step 6: Evaluate the New Limit

Simplify the expression \( \lim_{x \rightarrow 0} \frac{-\sin^2 x}{x \cdot \cos x} \), which reduces to \( \lim_{x \rightarrow 0} -x \cdot \tan x \). As \( x \rightarrow 0 \), the expression tends to 0.

Step 7: Conclude the Limit Calculation

Since the product approaches 0, \( \ln y \rightarrow 0 \) meaning \( y = |x|^{\sin x} \rightarrow 1 \) as \( e^0 = 1 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in Calculus used to find limits of indeterminate forms. Whenever you have a limit that initially appears to be in one of the indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule helps transcend the difficulty by differentiating the numerator and the denominator separately.
This rule states that under the right circumstances:

• If \( \lim_{x \rightarrow a} f(x) = 0 \) and \( \lim_{x \rightarrow a} g(x) = 0 \), or both approach infinity, then we can use the derivatives:
• \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \).

This process continues until the limit becomes determinate. In our exercise, we used L'Hôpital's Rule to transform the expression from \( \sin x \cdot \ln |x| \) to \( \frac{-\sin^2 x}{x \cdot \cos x} \), which ultimately simplifies to a form where the limit could be evaluated directly.

Limits in Calculus

Limits are foundational concepts in calculus that describe the behavior of a function as it approaches a particular point. Understanding limits allows mathematicians and students to discuss the tendencies of functions, particularly when they do not behave as intuitively expected.
For expression \( \lim_{x \to a} f(x) \), we seek the value that \( f(x) \) nears as \( x \) converges to \( a \).
Calculus uses limits to handle continuous functions, discontinuities, and more complex forms like indeterminate forms. In our problem, calculating the limit as \( x \to 0 \) required transformation techniques and applying L'Hôpital's Rule, highlighting how limits can sometimes lead to solutions through indirect paths.

Exponential Functions

Exponential functions involve expressions where a constant base is raised to a variable exponent, such as \( a^x \). In many mathematical problems, understanding the behavior of exponential functions helps evaluate growth processes, decay, and other varying phenomena.
In our exercise, the expression \( |x|^{\sin x} \) involved taking an absolute value raised to an oscillating exponent as \( x \to 0 \). Such situations call for transformations to linearize the exponential behavior, for example by using logarithms. By doing this, the multiplication becomes manageable and allows us to evaluate the limits through derivatives and eventually find that the expression simplifies to \( 1 \) due to the limit approaching its exponential form \( e^0 = 1 \).
By leveraging both limits and transformations, we unlock the behavior of seemingly complex functions quickly.