Problem 131 - A Complete Resource Book in Mathematics for JEE Main 2017

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: Recognize the Limit Formula

We need to find the limit \( \lim_{x \rightarrow 0} \frac{\log(3+x) - \log(3-x)}{x} \). Recognize this expression is structured similarly to a derivative using the definition of a derivative: \( \lim_{h \to 0} \frac{f(a+h) - f(a-h)}{2h} \). This limit resembles a derivative at a point.

Step 2: Use Logarithmic Properties

Apply the property of logarithms: \( \log a - \log b = \log\left(\frac{a}{b}\right) \). Therefore, the numerator \( \log(3+x) - \log(3-x) \) becomes \( \log\left(\frac{3+x}{3-x}\right) \).

Step 3: Rewrite the Expression

The limit now becomes \( \lim_{x \to 0} \frac{\log\left(\frac{3+x}{3-x}\right)}{x} \). We will need to simplify this further to evaluate the limit.

Step 4: Differentiate the Logarithmic Function

Recognize \( \frac{3+x}{3-x} \) as a function. Differentiate \( \log(u) \) with respect to \( x \), where \( u = \frac{3+x}{3-x} \). Using the chain rule, derive: \( \frac{d}{dx} \log(\frac{3+x}{3-x}) = \frac{1}{\frac{3+x}{3-x}} \cdot \frac{d}{dx}(\frac{3+x}{3-x}) \).

Step 5: Differentiate the Argument

Apply the quotient rule to differentiate the function \( \frac{3+x}{3-x} \). \( \frac{d}{dx}\left( \frac{3+x}{3-x} \right) = \frac{(3-x)(1) - (3+x)(-1)}{(3-x)^2} = \frac{3-x+3+x}{(3-x)^2} = \frac{6}{(3-x)^2} \).

Step 6: Simplify the Expression

Plug the derivative \( \frac{6}{(3-x)^2} \) back into the differentiated ln expression. Hence, \( \frac{d}{dx} \log\left(\frac{3+x}{3-x}\right) = \frac{1}{\frac{3+x}{3-x}} \cdot \frac{6}{(3-x)^2} = \frac{6}{(3+x)(3-x)} \). At \( x = 0 \), the expression simplifies to \( \frac{6}{9} = \frac{2}{3} \).

Step 7: Conclude the Limit

The derivative at \( x = 0 \) of the original function exactly equals the value of the limit \( \lim_{x \to 0} \frac{\log(3+x) - \log(3-x)}{x} \). Therefore, \( k = \frac{2}{3} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique in calculus used to differentiate functions that involve complicated products or quotients, as well as functions raised to a variable power. This method is particularly useful when dealing with logarithmic expressions, like the one found in our limit problem.

To use logarithmic differentiation, we first take the natural logarithm of both sides of the function we wish to differentiate. This process simplifies the function by turning multiplications into additions and divisions into subtractions through the logarithmic properties:

• Product Rule: \( \log(a \times b) = \log(a) + \log(b) \)
• Quotient Rule: \( \log(\frac{a}{b}) = \log(a) - \log(b) \)

In the original problem, we leveraged the property \( \log(3+x) - \log(3-x) = \log\left(\frac{3+x}{3-x}\right) \). This transformation simplified the limit we were working with, making it easier to differentiate.

After this simplification, deriving the function with respect to \( x \) can be more straightforward, as the derivative of \( \log(u) \) is \( \frac{1}{u} \frac{du}{dx} \), which leads us into using the chain rule.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the division of two other functions. It is especially handy when you can't simply apply the product rule. In mathematical terms, if you have a function \( f(x) = \frac{g(x)}{h(x)} \), the quotient rule formula is:

• \( \frac{d}{dx}\left(\frac{g(x)}{h(x)}\right) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \)

In our exercise, the expression \( \frac{3+x}{3-x} \) required the application of the quotient rule to find its derivative. Here:

• \( g(x) = 3+x \)
• \( h(x) = 3-x \)

By applying the quotient rule, we computed the derivative:\[\frac{d}{dx}\left(\frac{3+x}{3-x}\right) = \frac{(3-x)(1) - (3+x)(-1)}{(3-x)^2} = \frac{6}{(3-x)^2}\]This derivative is then used in combination with logarithmic differentiation to solve the limit problem at hand.

Chain Rule

The chain rule is an essential rule in calculus used to differentiate composite functions. When you have a function inside another function, the chain rule allows you to find the derivative of this composite function efficiently. It states:

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

In our exercise, after rewriting the original limit expression using logarithmic properties, we arrived at a composite function: \( \log(\frac{3+x}{3-x}) \). This is where the chain rule was applied.

We identified the inner function \( u = \frac{3+x}{3-x} \) and the outer function \( v = \log(u) \). Differentiating \( \log(u) \) with respect to \( x \) involves multiplying the derivative of the outer function \( \frac{1}{u} \) by the derivative of the inner function \( \frac{du}{dx} \), which was obtained via the quotient rule.

The chain rule linked these two steps and simplified our expression to:\[\frac{d}{dx} \log\left(\frac{3+x}{3-x}\right) = \frac{1}{\frac{3+x}{3-x}} \cdot \frac{6}{(3-x)^2} = \frac{6}{(3+x)(3-x)}\]Finally, evaluating this expression at \( x = 0 \) provided our limit, confirming the value of \( k \) as \( \frac{2}{3} \).