Problem 131 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\lim _{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=k\), the value of \(k\) is (A) 0 (B) \(-\frac{1}{3}\) (C) \(\frac{2}{3}\) (D) \(-\frac{2}{3}\)

Step-by-Step Solution

Answer

The value of \( k \) is \( \frac{2}{3} \), option (C).

Step 1: Apply L'Hôpital's Rule

The limit is in the indeterminate form \( \frac{0}{0} \) as \( x \to 0 \). To solve it, apply L'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the limit yields an indeterminate form. Differentiate the numerator and the denominator with respect to \( x \).

Step 2: Differentiate the Numerator

Differentiate the expression \( \log(3+x) - \log(3-x) \) with respect to \( x \). The derivative is \( \frac{1}{3+x} + \frac{1}{3-x} \) since the derivative of \( \log(u) \) with respect to \( x \) is \( \frac{1}{u} \frac{du}{dx} \).

Step 3: Differentiate the Denominator

The derivative of the denominator \( x \) with respect to \( x \) is \( 1 \).

Step 4: Express the New Limit Form

Now substitute the derivatives into the limit expression to find: \[ \lim_{x \to 0} \frac{\frac{1}{3+x} + \frac{1}{3-x}}{1} = \frac{1}{3+x} + \frac{1}{3-x}. \]

Step 5: Simplify and Evaluate the Limit

Simplify \( \frac{1}{3+x} + \frac{1}{3-x} \) and evaluate it as \( x \to 0 \): \[ \lim_{x \to 0} \left( \frac{1}{3+x} + \frac{1}{3-x} \right) = \frac{1}{3+0} + \frac{1}{3-0} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]

Step 6: Conclusion

The value of the limit is \( k = \frac{2}{3} \). Therefore, the answer to the problem is option (C).

Key Concepts

Indeterminate Forms

Indeterminate forms can be confusing at first, but they are an important part of limit evaluation. These forms typically arise in calculus when you try to directly substitute a value into a function and get an undefined or ambiguous answer.
Indeterminate forms include expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0\cdot\infty \), \( \infty - \infty \), to name a few. In many cases, these forms suggest that the value of the limit is not immediately apparent and further analysis is needed.

• The form \( \frac{0}{0} \) is common in limit problems. It suggests that both the numerator and denominator approach zero, making the limit seemingly undefined.
• In the given exercise, the initial expression \( \frac{\log(3+x) - \log(3-x)}{x} \) resulted in the indeterminate form \( \frac{0}{0} \) as \( x \to 0 \).
• To resolve this, we use a powerful tool called L'Hôpital's Rule, which applies to certain indeterminate forms to help find the limit.

By understanding these forms, and knowing how to apply mathematical techniques to resolve them, you're better equipped to tackle calculus problems involving limits.

Logarithmic Differentiation

Logarithmic differentiation is a handy technique, especially useful when differentiating expressions involving logarithms. It helps simplify the differentiation process in cases like products or quotients.
In the exercise, we needed to differentiate \( \log(3+x) - \log(3-x) \) with respect to \( x \) to apply L'Hôpital's Rule.

• For a logarithmic function \( \log(u) \), its derivative is \( \frac{1}{u} \frac{du}{dx} \), by the chain rule.
• For \( \log(3+x) \), the derivative becomes \( \frac{1}{3+x} \) since the derivative of \( 3+x \) is 1.
• Similarly, the derivative of \( \log(3-x) \) becomes \(-\frac{1}{3-x} \) because the derivative of \( 3-x \) is -1.

Using logarithmic differentiation, the process becomes more straightforward, allowing us to break down complex expressions step-by-step. This leads to more precise and manageable calculations when dealing with limits and other calculus operations.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus, allowing us to understand the behavior of functions as they approach specific points.
In the context of this exercise, we were tasked with finding the limit of \( \frac{\log(3+x) - \log(3-x)}{x} \) as \( x \to 0 \).

• The initial approach failed as it resulted in the indeterminate form \( \frac{0}{0} \).
• By applying L'Hôpital's Rule, we differentiated the numerator and the denominator separately. This transformed the problem into a form we could solve.
• We ended up with the expression \( \frac{1}{3+x} + \frac{1}{3-x} \).

After substituting \( x = 0 \), the expression simplified to \( \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \) which is the value of the limit \( k \).
Limit evaluation techniques, including L'Hôpital's Rule and differentiation strategies, are crucial tools when dealing with indeterminate forms. They enable us to precisely analyze the behavior of functions in calculus.