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

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 \(\frac{2}{9}\); answer (A).

Step 1: Understand the Expression

We need to find the limit \( \lim _{x \rightarrow 0} \frac{\ln (2-\cos 2x)}{\ln ^{2}(\sin 3x+1)} \). This expression involves natural logarithms and trigonometric functions, which must be handled carefully as \(x\) approaches 0.

Step 2: Expand Cosine and Sine Functions

As \(x\) approaches 0, we utilize Taylor series expansions: \(\cos 2x \approx 1 - 2x^2\) and \(\sin 3x \approx 3x\). Substitute these approximations into the expressions.

Step 3: Simplify the Numerator

Replace \(\cos 2x\) with its approximation to get \(\ln(2-\cos 2x) = \ln(1 + 2x^2) \). Using \(\ln(1 + u) \approx u\) for small \(u\), simplify to \(2x^2\).

Step 4: Simplify the Denominator

Replace \(\sin 3x\) with its approximation to get \(\ln^2(\sin 3x + 1) = \ln^2(1+3x)\). Using the expansion \(\ln(1 + v) \approx v\), simplify to \((3x)^2 = 9x^2\).

Step 5: Evaluate the Limit

Substitute the simplified forms from steps 3 and 4 back into the original expression: \(\frac{2x^2}{9x^2}\). Cancel the \(x^2\) terms to obtain \(\frac{2}{9}\).

Step 6: Conclusion

The expression simplifies to a constant value \(\frac{2}{9}\) as \(x\) approaches 0. Check the options to find that the correct answer is (A).

Key Concepts

Trigonometric Functions

Trigonometric functions are critical components of calculus, especially when evaluating limits. These functions, namely sine and cosine, vary periodically and are defined using the unit circle. In the context of limits, understanding their behavior as they approach particular points is essential. For the problem at hand, we focus on small angles, and understanding the linear approximations of trigonometric functions is crucial. As a rule of thumb,

• \( \sin x \approx x \) when \( x \) is close to 0.
• \( \cos x \approx 1 - \frac{x^2}{2} \) under similar conditions.

These approximations, derived from Taylor series expansions, simplify calculations significantly in limits. Therefore, getting comfortable with these expansions helps solve such limit problems efficiently.

Natural Logarithms

Natural logarithms are logarithms with a base of \( e \), a fundamental constant approximately equal to 2.718. They are prevalent in many fields like mathematics, physics, and engineering due to their properties and natural growth models. In calculus, natural logarithms help simplify complex expressions, especially when dealing with small numbers close to 1. For the exercise involving limits, the formula \( \ln(1 + u) \approx u \), where \( u \) is a small number, becomes extremely useful. This approximation allows us to break down the logarithmic functions in the numerator and denominator into simpler terms, enabling us to compute the limit efficiently.

• Remember, when dealing with \( \ln(1 + x) \), if \( x \) is very small, you can approximate it as \( x \).

This insight is often tested in limit problems, so familiarity with these properties can streamline the solving process.

Taylor Series Expansion

The Taylor series expansion is a powerful tool in calculus for expressing functions as infinite sums of terms calculated from the values of their derivatives at a single point. It allows us to approximate complicated functions with simpler polynomials, which is especially useful when dealing with limits and small inputs. In our specific exercise, the Taylor series expansion helps transform trigonometric functions to their approximations:

• \( \sin x \) around 0 becomes \( x - \frac{x^3}{6} + \cdots \).
• \( \cos x \) becomes \( 1 - \frac{x^2}{2} + \cdots \).

These expressions simplify calculations by reducing complex functions to their linear terms. It's important to recognize which terms to keep and which to discard when performing these expansions in various calculus problems. Understanding this concept makes manipulating functions for limits, derivatives, and integrals much more manageable.

Calculus Problem Solving

Solving calculus problems, particularly limits, involves a mix of algebraic manipulation and a deep understanding of mathematical principles. The exercise you encountered demonstrates a typical calculus problem-solving path: break down the problem into simpler components using known approximations. Key steps include:

• Identifying applicable expansions for relevant functions, such as trigonometric functions and logarithms.
• Substituting these approximations into the problem context.
• Performing algebraic simplification, ensuring each step is consistent with assumptions made by approximations.

Finally, conclude by evaluating the limit of the simplified expression. Mastery in these kinds of calculus problems comes with practice and understanding foundational concepts like Taylor series and logarithmic behavior. These skills not only solve textbook problems but also apply widely in real-world scenarios involving calculus.