Problem 53 - Calculus - AP Edition

Question

Find the limit by interpreting the expression as an appropriate derivative. (a) \(\lim _{x \rightarrow 0} \frac{\ln (1+3 x)}{x}\) (b) \(\lim _{x \rightarrow 0} \frac{\ln (1-5 x)}{x}\)

Step-by-Step Solution

Answer

(a) 3, (b) -5

Step 1: Identify the Limit as a Derivative

The limit \( \lim_{x \rightarrow 0} \frac{\ln(1+3x)}{x} \) resembles the definition of a derivative \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} \). Here, interpret the expression as the derivative of \( f(x) = \ln(1+3x) \) at \( x = 0 \).

Step 2: Apply the Derivative Definition

Since \( f(x) = \ln(1+3x) \), find the derivative \( f'(x) \) using the chain rule. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \); thus, \( f'(x) = \frac{1}{1+3x} \cdot 3 \).

Step 3: Evaluate the Derivative at the Point

Calculate \( f'(0) \), which is \( \frac{3}{1+3 \times 0} = 3 \). Therefore, the limit \( \lim_{x \rightarrow 0} \frac{\ln(1+3x)}{x} \) is 3.

Step 4: Apply Same Concept for Second Limit

Recognize \( \lim_{x \rightarrow 0} \frac{\ln(1-5x)}{x} \) as the derivative \( f'(x) \) of \( f(x) = \ln(1-5x) \) at \( x = 0 \).

Step 5: Find the Derivative Using the Chain Rule

The derivative of \( f(x) = \ln(1-5x) \) is \( f'(x) = \frac{1}{1-5x} \cdot (-5) = \frac{-5}{1-5x} \).

Step 6: Evaluate the Derivative at X=0

Calculate \( f'(0) \), which is \( \frac{-5}{1-0} = -5 \). Thus, the limit \( \lim_{x \rightarrow 0} \frac{\ln(1-5x)}{x} \) is -5.

Key Concepts

Derivative Interpretation

In calculus, understanding the concept of a derivative is essential. When we look at an expression like \ \ \( \lim _{x \rightarrow 0} \frac{\ln (1+3 x)}{x} \), we identify it as resembling the definition of a derivative. In general, the derivative of a function \( f(x) \) at a point \( a \) is given by:

• \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} \)

This gives a measure of how the function \( f(x) \) changes around the point \( a \).
The key is recognizing the expression within the limit as a derivative. The numerator here, \( \ln (1+3x) \), suggests that this is the derivative of \( \ln \) evaluated at a point close to zero. Understanding derivative interpretation helps break down complex limit problems efficiently.

Natural Logarithm

The natural logarithm, denoted by \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It's a fundamental concept in calculus and has particular properties that simplify differentiation processes.
The rule to remember about natural logarithms is that the derivative of \( \ln(u) \) with respect to \( u \) is:

• \( \frac{d}{du} [\ln(u)] = \frac{1}{u} \)

This property makes derivative operations straightforward when dealing with logarithmic functions.
In our problem, understanding this property is key, as it directly affects how the derivative calculation proceeds when applying the chain rule in evaluating the limit. The natural logarithm's base \( e \) properties lead to some of the simplest forms of derivatives.

Chain Rule

The chain rule is a critical tool in calculus for finding the derivative of composite functions. Essentially, it states:

• If you have a composite function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).

In the given exercise, the chain rule helps in differentiating expressions like \( \ln(1+3x) \) and \( \ln(1-5x) \). Here are the steps to apply the chain rule:

• First, differentiate with respect to the inner function, \( u=1+3x \), or \( u=1-5x \).
• Secondly, multiply by the derivative of the inner function, \( u \), which gives us the complete derivative.

The chain rule shows how each function inside another "chains along," contributing to the overall derivative. It's an effective strategy for breaking down and assessing each part separately.

Limit Evaluation

Evaluating limits is a core technique in calculus, especially in contexts such as determining instantaneous rates of change or finding derivatives. In our example, evaluating limits provides insight into the behavior of logarithmic functions as they approach certain points.
Consider limits of forms such as \( \lim_{x \rightarrow 0} \frac{\ln(1+3x)}{x} \). By recognizing this expression as a derivative, limit evaluation becomes much more straightforward. Here are some steps:

• Identify if the expression can be seen as a derivative. In our problem, it aligns with the derivative form at \( x = 0 \).
• Use algebraic manipulation or known derivative results to simplify the evaluation.

By interpreting the limit as a derivative, as we've done in the provided exercise, we discover the behavior of complex functions as they approach particular values.