Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 0} \frac{\ln (2-\cos 2 x)}{\ln ^{2}(\sin 3 x+1)}\) is equal to (A) \(\frac{2}{9}\) (B) \(-\frac{2}{9}\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The limit is 0. (Option C)

Step 1: Analyze the Limit Expression

We need to evaluate the limit \( \lim_{x \rightarrow 0} \frac{\ln (2-\cos 2x)}{\ln^2(\sin 3x + 1)} \). Both the numerator \( \ln(2-\cos 2x) \) and the denominator \( \ln^2(\sin 3x+1) \) become \( \ln(1) = 0 \) as \( x \rightarrow 0 \), which results in an indeterminate form \( \frac{0}{0} \). This suggests using l'Hopital's rule.

Step 2: Apply l'Hopital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we apply l'Hopital's rule, which requires us to differentiate the numerator and the denominator with respect to \( x \).

Step 3: Differentiate the Numerator

The numerator is \( \ln(2-\cos 2x) \). Using the chain rule, the derivative is \( \frac{d}{dx} \big[\ln(2-\cos 2x)\big] = \frac{1}{2-\cos 2x} \cdot (\sin 2x) \cdot 2 = \frac{2\sin 2x}{2-\cos 2x} \).

Step 4: Differentiate the Denominator

The denominator is \( \ln^2(\sin 3x + 1) \). Using the chain rule, the derivative is \( 2\ln(\sin 3x + 1) \cdot \frac{1}{\sin 3x + 1} \cdot (\cos 3x \cdot 3) = \frac{6\cos 3x \cdot \ln(\sin 3x + 1)}{\sin 3x + 1} \).

Step 5: Evaluate the New Limit Expression

The expression becomes \( \lim_{x \rightarrow 0} \frac{\frac{2\sin 2x}{2-\cos 2x}}{\frac{6\cos 3x \cdot \ln(\sin 3x + 1)}{\sin 3x + 1}} = \lim_{x \rightarrow 0} \frac{2\sin 2x (\sin 3x + 1)}{6\cos 3x \cdot \ln(\sin 3x + 1)(2-\cos 2x)} \). As \( x \rightarrow 0 \), \( \sin kx \approx kx \) for small \( x \) and \( 2-\cos 2x \approx 2 - (1 - 2x^2) \approx 2x^2 \).

Step 6: Simplify the Expression and Final Evaluation

Approximate when \( x \rightarrow 0 \): \( \sin 2x \approx 2x \), \( \sin 3x \approx 3x \), and using basic properties, approximate \( \ln(1+u) \approx u \), thus \( \ln(\sin 3x + 1) \approx 3x \). Plug these into the expression:\[ \lim_{x \rightarrow 0} \frac{2(2x)(3x+1)}{6(1) \cdot 3x \cdot 2x^2} \approx \lim_{x \rightarrow 0} \frac{12x^2 + 2x}{36x^3} \] Simplify:\[ = \lim_{x \rightarrow 0} \frac{12x + 2}{36x^2} = \lim_{x \rightarrow 0} \frac{12}{36x} + \frac{2}{36x^2} = 0 \] Conclusion: The answer is 0.

Key Concepts

Indeterminate forms in Limits

In calculus, when evaluating limits, you often encounter expressions that are not straightforward. They form types that give unclear or undefined outcomes, known as indeterminate forms. One common indeterminate form is \( \frac{0}{0} \), which occurs when both the numerator and the denominator of a rational expression approach zero.
In our exercise, as \(x \rightarrow 0\), both \( \ln (2 - \cos 2x) \) and \( \ln^2(\sin 3x + 1) \) tend towards zero, resulting in the indeterminate form \( \frac{0}{0} \).
To solve these types of problems, we rely on more advanced methods such as l'Hopital's rule to evaluate the limit and better understand the behavior of the function near the point of indeterminacy. Remember, an indeterminate form indicates that further manipulation or simplification is necessary to find the true limit.

Using l'Hopital's Rule

l'Hopital's rule is a powerful tool in calculus used to resolve limits that present indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) presents such a form, and both \( f(x) \) and \( g(x) \) are differentiable, then:

• \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \),

provided that the limit on the right side exists.
In the original exercise, the function was indeterminate in the form \( \frac{0}{0} \). Therefore, we differentiate the numerator \( \ln(2- \cos 2x) \) and the denominator \( \ln^2(\sin 3x + 1) \) using the rules of differentiation to find their derivatives.
Once differentiated, we re-evaluate the limit with these new expressions. l'Hopital's rule simplifies finding the limit by transforming a complex form into a simpler one. Always ensure derivatives are correctly calculated to apply the rule effectively.

The Chain Rule in Differentiation

The chain rule is a fundamental derivative rule in calculus, particularly useful when dealing with composite functions. A composite function is one that applies multiple functions to a variable, for example, \( f(g(x)) \).
The chain rule states:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This means you first take the derivative of the outer function and multiply it by the derivative of the inner function.
In our exercise, when differentiating \( \ln(2 - \cos 2x) \) and \( \ln^2(\sin 3x + 1) \), we applied the chain rule. For the first, the derivative is \( \frac{1}{2-\cos 2x} \cdot (-\sin 2x \cdot 2) \), and similarly for the second, we used the chain rule to differentiate the composite function.
Understanding and applying the chain rule is crucial, especially when performing operations like l'Hopital's rule, where differentiation of complex composite functions often occurs.