Problem 82 - Calculus - AP Edition

Question

Find the limit by interpreting the expression as an appropriate derivative. $$\lim _{w \rightarrow 2} \frac{3 \sec ^{-1} w-\pi}{w-2}$$

Step-by-Step Solution

Answer

The limit is \( \frac{\sqrt{3}}{2} \).

Step 1: Identify a Derivative Form

The given limit expression \( \lim _{w \rightarrow 2} \frac{3 \sec^{-1} w - \pi}{w-2} \) resembles the definition of a derivative at a point. If we consider \( f(w) = 3 \sec^{-1} w \), the expression takes the form of \( \lim _{w \rightarrow a} \frac{f(w) - f(a)}{w-a} \) where \( a = 2 \).

Step 2: Define the Function and Derivative Point

We recognize \( f(w) = 3 \sec^{-1} w \) and evaluate the derivative at \( w = 2 \). The point \( a \) is 2, and we need \( f'(w) \), the derivative of \( f(w) = 3 \sec^{-1} w \).

Step 3: Differentiate the Function

The derivative of \( \sec^{-1} x \) is \( \frac{1}{x \sqrt{x^2 - 1}} \). Therefore, the derivative of \( f(w) = 3 \sec^{-1} w \) is \( f'(w) = 3 \times \frac{1}{w \sqrt{w^2 - 1}} = \frac{3}{w \sqrt{w^2 - 1}} \).

Step 4: Evaluate the Derivative at the Point

Substitute \( w = 2 \) into \( f'(w) \) to evaluate the derivative at this point. So, \( f'(2) = \frac{3}{2 \sqrt{2^2 - 1}} = \frac{3}{2 \sqrt{3}} \).

Step 5: Simplify the Expression

Simplify \( \frac{3}{2 \sqrt{3}} \) by rationalizing the denominator: \( \frac{3}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \).

Key Concepts

Calculus Derivatives

Calculus derivatives let us find the rate at which a function is changing at any given point. Think of it as asking how fast a car is going right now, rather than its average speed over a whole journey. In this exercise, we see a limit expression, which represents a derivative. The expression \( \lim _{w \rightarrow 2} \frac{3 \sec^{-1} w - \pi}{w-2} \) looks similar to a formula for finding the derivative of a function: \( \lim _{w \rightarrow a} \frac{f(w) - f(a)}{w-a} \). This specific setup corresponds to evaluating the derivative of the function \( f(w) = 3 \sec^{-1} w \) at the point \( w = 2 \). By finding that derivative, we determine how \( f(w) \) changes as \( w \) changes very close to 2. This involves using knowledge of basic derivative rules.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles when given trigonometric ratios. The inverse secant function, \( \sec^{-1} x \), helps us find an angle \( \theta \) such that \( \sec \theta = x \). In the function \( f(w) = 3 \sec^{-1} w \) from our problem, \( \sec^{-1} \) allows us to evaluate changes in angular measure with respect to the input \( w \). The derivation of \( \sec^{-1} x \) involves certain complexities because the domain and range are adjusted compared to linear functions.
For differentiation, which is our task, we use the known derivative: \( \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x \sqrt{x^2 - 1}} \). Applying this rule makes it possible to determine how the function \( f(w) = 3 \sec^{-1} w \) changes as \( w \) is altered slightly—specifically at the point \( w = 2 \), leading us to the derivative \( f'(w) = \frac{3}{w \sqrt{w^2 - 1}} \).

Rationalizing the Denominator

Rationalizing the denominator is a technique used to remove any irrational numbers (like square roots) from the bottom (denominator) of a fraction. This is often done to make expressions look simpler and is a common step in problems involving limits and derivatives.
In our problem, once we find \( f'(2) = \frac{3}{2 \sqrt{3}} \), we use rationalization to simplify this result. Here’s how it works:

• We multiply the numerator and the denominator by \( \sqrt{3} \) to eliminate the square root from the denominator.
• This gives us \( \frac{3 \sqrt{3}}{2 \times 3} \), simplifying to \( \frac{\sqrt{3}}{2} \).

This step is important not only for tidying up the final answer but also for practices where we need a rational number in the denominator for further calculations or interpretations.