Problem 192 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{\ln (5+x)-\ln (5-x)}{x} .\left\\{\text { Ans. } \frac{2}{5}\right\\} $$

Step-by-Step Solution

Answer

The given expression is \(\lim_{x\rightarrow0} \frac{\ln(5+x)-\ln(5-x)}{x}\). We can simplify it using logarithm properties to get \(\lim_{x\rightarrow0} \frac{\ln(\frac{5+x}{5-x})}{x}\). Applying L'Hopital's Rule, we differentiate the numerator and denominator with respect to x, yielding \(\lim_{x\rightarrow0} \frac{\frac{10}{(5+x)(5-x)^{2}}}{1}\). Evaluating this limit as x approaches 0 gives us \(\frac{2}{5}\).

Step 1: Simplify the expression using logarithm properties

The given expression is \(\frac{\ln(5+x)-\ln(5-x)}{x}\). As the properties of logarithms states that: \[ \ln a - \ln b = \ln \left(\frac{a}{b}\right) \] So we can simplify the expression to: \[ \lim_{x\rightarrow0} \frac{\ln(\frac{5+x}{5-x})}{x} \]

Step 2: Apply L'Hopital's Rule

Since the limit is in the indeterminate form of \(\frac{0}{0}\), we can apply L'Hopital's Rule. Differentiate both the numerator and the denominator by x: \[ \lim_{x\rightarrow0} \frac{\frac{d}{dx}\ln(\frac{5+x}{5-x})}{\frac{d}{dx}x} \] The derivative of the numerator is done using the chain rule: \[ \frac{d}{dx} \ln(\frac{5+x}{5-x}) = \frac{10}{(5+x)(5-x)^{2}} \] The derivative of x is 1. Therefore, the new expression becomes: \[ \lim_{x\rightarrow0} \frac{\frac{10}{(5+x)(5-x)^{2}}}{1} \]

Step 3: Evaluate the limit

Now we can substitute x=0 into the expression: \[ \frac{10}{(5+0)(5-0)^{2}} \] This simplifies to: \[ \frac{10}{5 \cdot 25} = \frac{2}{5} \] So, the limit as x approaches 0 is: \[ \lim _{x \rightarrow 0} \frac{\ln (5+x)-\ln (5-x)}{x} = \frac{2}{5} \]

Key Concepts

Logarithm Properties

Logarithm properties are incredibly useful for simplifying expressions, especially in calculus. One key property is the difference rule: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This means that the difference of two logarithms can be written as the logarithm of a quotient.
In the original exercise, this property helps to simplify \( \ln(5+x) - \ln(5-x) \) to \( \ln\left(\frac{5+x}{5-x}\right) \). This simplification is crucial for applying further calculus techniques.

• Allows simplification of complex logarithmic expressions.
• Makes other mathematical operations easier to perform.

Understanding these properties can greatly enhance your efficiency when working with limits involving logarithms.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. When you have a function inside another function, the chain rule helps you find the derivative of this composite function.
In our problem, we need to find the derivative of \( \ln\left(\frac{5+x}{5-x}\right) \). This involves not just differentiating \( \ln(u) \), but also the inner function \( \frac{5+x}{5-x} \).
The chain rule states:

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

In our case, this requires differentiating the inner function \( \frac{5+x}{5-x} \) and then multiplying by the derivative of the natural log function. This results in:
\[ \frac{d}{dx} \ln\left(\frac{5+x}{5-x}\right) = \frac{10}{(5+x)(5-x)^{2}}. \] Understanding the chain rule is crucial for tackling complex differentiation tasks.

Indeterminate Form

Indeterminate forms often occur in calculus when evaluating limits, especially when both the numerator and denominator approach zero. These forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others, and they often require special techniques like L'Hopital's Rule to resolve.
In the problem at hand, the limit expression initially takes the \( \frac{0}{0} \) form as \( x \rightarrow 0 \). This specific indeterminate form signals that simple substitution is not enough, and more analysis is needed.
L'Hopital's Rule is a method to resolve this by differentiating the numerator and the denominator separately, then retrying the limit:

• If \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \), then \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) may be used.

This powerful rule allows continuation even when direct substitution leads to an undefined state, helping find exact values for limits you might encounter.