Problem 105 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\lim _{x \rightarrow 0} \frac{x^{a} \sin ^{b} x}{\sin \left(x^{c}\right)}, a, b, c \in R-\\{0\\}\) exists and has non-zero value, then (A) \(a, b, c\) are in A.P. (B) \(a, c, b\) are in A.P. (C) \(a, c, b\) are in G.P. (D) None of these

Step-by-Step Solution

Answer

Option A: \( a, b, c \) are in A.P.

Step 1: Analyze the limit

We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{x^{a} \sin ^{b} x}{\sin(x^{c})} \). First, apply L'Hôpital's Rule as directly plugging in \(x = 0\) gives an indeterminate form 0/0. Since both the numerator and the denominator approach zero, L'Hôpital's Rule is applicable.

Step 2: Differentiate numerator and denominator

Differentiate the numerator: \( f(x) = x^a \sin^b(x) \). Using the product rule, we get the derivative \( f'(x) = a x^{a-1} \sin^b(x) + b x^a \sin^{b-1}(x) \cos(x) \). Differentiate the denominator: \( g(x) = \sin(x^c) \). Using the chain rule, the derivative is \( g'(x) = c x^{c-1} \cos(x^c) \).

Step 3: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, the limit becomes:\[ \lim _{x \rightarrow 0} \frac{a x^{a-1} \sin^b(x) + b x^a \sin^{b-1}(x) \cos(x)}{c x^{c-1} \cos(x^c)}. \]Now, factor out the terms which will be helpful in understanding the behavior of the limit as x approaches 0.

Step 4: Factor and Simplify the Limit

Factor out the terms that can be separated from the limit analysis:\[ \lim_{x \to 0} \frac{x^{a-1} \sin^b(x)(a + b x \cot(x))}{c x^{c-1} \cos(x^c)}. \]As \(x \) approaches 0, \(\sin(x) \sim x\) and \(\cos(x^c) \sim 1\). Simplifies to\[ \lim_{x \to 0} \frac{x^{a + b -1}}{x^{c-1}} \times \frac{a + b x \cot(x)}{c}. \]

Step 5: Determine Condition for Non-Zero Limit

For the expression \(\frac{x^{a + b - 1}}{x^{c-1}} \) to have a non-zero limit, equate the powers of x,\[ a + b - 1 = c - 1 \Rightarrow a + b = c. \]

Step 6: Analyze the Sequence Type

The equation \( a + b = c \) suggests that \( a, b, c \) are in arithmetic progression since the middle term \( b \) is the mean of \( a \) and \( c \) showing that \( 2b = a + c \).

Key Concepts

L'Hôpital's Rule

In calculus, L'Hôpital's Rule helps solve limits where both the numerator and the denominator approach zero or infinity, creating an indeterminate form. When we work with these tricky situations, the rule provides a straightforward way to find the limit.

Given two functions, say \( f(x) \) and \( g(x) \), both approaching zero or infinity as \( x \) approaches some value, L'Hôpital's Rule states that:

• If \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \) or \( \pm \infty \),
• Then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this latter limit exists.

This is particularly useful as it allows us to deal directly with derivatives, making the computation of limits more manageable. As seen in the step-by-step solution, applying L'Hôpital's Rule involves differentiating both the numerator and the denominator and then re-evaluating the limit.

Arithmetic Progression (A.P.)

An arithmetic progression (A.P.) is a sequence of numbers in which the difference of any two successive members is a constant, called the common difference. This simple property makes arithmetic progressions predictable and easy to work with.

Consider an A.P. characterized by:

• a first term \( a_1 \),
• and a common difference \( d \).

The general form of the A.P. will be \( a_1, a_1 + d, a_1 + 2d, \ldots \). Each term in the series can be expressed as \( a_n = a_1 + (n-1)d \).

In the context of our exercise, we note that if \( a, b, c \) meet the A.P. condition \( a + b = c \), then it implies a property where each number is related through a consistent step or difference, reinforcing their sequence as an arithmetic progression.

Indeterminate Forms

Indeterminate forms in calculus are expressions that do not have a clear, defined limit. These forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others. They emerge often in limits and require special techniques like L'Hôpital's Rule to resolve.

When initially examining a limit problem, such as in the provided exercise, plugging \( x = 0 \) into \( \frac{x^a \sin^b(x)}{\sin(x^c)} \) produces a \( \frac{0}{0} \) form, indicating an indeterminate form. Solving this form is only possible when reworking the expression, finding a pattern or as seen in L'Hôpital's application, differentiating the separate components.

Understanding these indeterminate forms is crucial because what appears as undefined at first glance can actually simplify to a finite value or even reveal a specific sequence type, as shown in the exercise where the limit transformed to indicate an arithmetic progression.