Problem 212 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{e^{a x}-e^{-2 a x}}{\ln (1+x)}\\{\text { Ans. } 3 a\\} $$

Step-by-Step Solution

Answer

The short answer for the given limit question, using L'Hôpital's Rule and simplifying the expression, is \(3a\).

Step 1: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as x approaches 0, we can apply L'Hôpital's Rule. To do this, we need to compute the derivatives of both the numerator and denominator with respect to x and then take the limit of the ratio of the derivatives. The derivative of the given function's numerator is \( \frac{d}{dx}(e^{ax}-e^{-2ax}) \) and the derivative of the denominator is \( \frac{d}{dx}(\ln(1+x)) \).

Step 2: Compute the derivatives

Apply the chain rule to compute the derivatives: For the numerator: \( \frac{d}{dx}(e^{ax}-e^{-2ax}) = ae^{ax} + 2ae^{-2ax} \). For the denominator: \( \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} \).

Step 3: Compute the limit of the ratio of the derivatives

As we apply L'Hôpital's Rule, our equation becomes: \( \lim_{x \rightarrow 0} \frac{ae^{ax} + 2ae^{-2ax}}{\frac{1}{1+x}} \). Now, we will compute the limit by directly substituting x=0 into the equation (since this substitution is valid): \( \lim_{x \rightarrow 0} \frac{ae^{ax} + 2ae^{-2ax}}{\frac{1}{1+x}} = \frac{ae^{a(0)} + 2ae^{-2a(0)}}{\frac{1}{1+0}} \).

Step 4: Simplify the expression

Simplify the Expression: \( \frac{ae^{0} + 2ae^{0}}{1} = a(1) + 2a(1) = 3a \). The solution to the limit is 3a.

Key Concepts

Chain Rule

The Chain Rule is an essential concept in calculus used to differentiate composite functions. It gives us a method to compute the derivative of a function that is nested inside another function. Suppose we have two functions, say \( u(x) \) and \( v(x) \), and a composite function \( f(x) = u(v(x)) \). To find the derivative of \( f \), we use the Chain Rule:

• Take the derivative of the outer function \( u(v) \) with respect to \( v \).
• Multiply it by the derivative of the inner function \( v(x) \).

In our exercise, the Chain Rule helps to find the derivative of the exponential functions \( e^{ax} \) and \( e^{-2ax} \). Differentiating \( e^{ax} \) gives \( ae^{ax} \), while differentiating \( e^{-2ax} \) results in \(-2ae^{-2ax} \). This rule is crucial for handling functions with complex layers, providing a straightforward path to its differentiation.

Derivatives

Derivatives measure how a function changes as its input changes. They are immensely useful in calculus for understanding the rate of change and slopes of curves. Given a function \( f(x) \), the derivative \( f'(x) \) tells us how \( f \) changes with a small change in \( x \).
For our problem, we are interested in the derivatives of both the numerator and the denominator. By computing \( \frac{d}{dx}(e^{ax}-e^{-2ax}) = ae^{ax} + 2ae^{-2ax} \), we find the rate of change of the function in the numerator. For the denominator \( \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} \), this derivative gives the slope of the natural logarithm function that changes near 0.

• Differentiation gives insights into the function's behavior.
• It's a foundational tool for further calculus calculations such as limits and optimizing functions.

Exponential Functions

Exponential functions are widely studied in mathematics due to their unique properties. These functions are characterized by a constant base raised to a variable exponent, usually taking the form \( e^{x} \), where \( e \) is Euler's number (approximately 2.71828). These functions grow rapidly and appear frequently in natural processes such as population growth and radioactive decay.
In our particular problem, exponential functions \( e^{ax} \) and \( e^{-2ax} \) are present in the numerator of our ratio. Their derivatives, as calculated using the Chain Rule, contribute significantly to evaluating the limit problem using L'Hôpital's Rule. These derivatives are \( ae^{ax} \) for \( e^{ax} \), reflecting how fast the function grows/shrinks depending on \( a \).
Exponential derivatives showcase growth patterns that change swiftly, underlying their significance in calculus and real-world applications.

Limits

Limits are a fundamental concept in calculus, providing a way to determine what value a function approaches as its input gets closer to a certain point. In the context of the given exercise, we compute the limit for the ratio of derivatives as \( x \) approaches 0 by applying L'Hôpital's Rule. This rule allows us to handle indeterminate forms like \( \frac{0}{0} \) that arise in our original problem.
After finding the derivatives, we substitute \( x = 0 \) into the derived expression to find:

• \( \lim_{x \rightarrow 0} \frac{ae^{ax} + 2ae^{-2ax}}{\frac{1}{1+x}} = 3a \)

This value gives us the solution, indicating how changing \( x \) affects the originally indeterminate form. Understanding limits enables us to navigate through problems where direct substitution doesn't initially work, showcasing their importance in calculus.