Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow(\pi / 2)^{-}}(\cos 2 x)^{x-\pi / 2}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Confirm the Form

To use l'Hôpital's Rule, we need to ensure the expression is in an indeterminate form. When evaluating \( x \to (\pi/2)^- \), \( \cos(2x) \to 0 \) from the positive side, and \( x - \pi/2 \to 0^- \). Thus, we have the form \( 0^0 \), which is an indeterminate form suitable for l'Hôpital's Rule.

Step 2: Use Logarithms to Simplify

To handle the form \( 0^0 \), start by taking the natural logarithm: \( \ln L = \lim_{x \to (\pi/2)^-} (x - \pi/2) \ln(\cos(2x)) \). This transforms the problem into \( \lim_{x \to (\pi/2)^-} f(x)/g(x) \) where \( f(x) = \ln(\cos(2x)) \) and \( g(x) = 1/(x-\pi/2) \).

Step 3: Find Derivatives for l'Hôpital's Rule

Differentiate \( f(x) \) and \( g(x) \) with respect to \( x \): - \( f'(x) = \frac{-2\sin(2x)}{\cos(2x)} \) - \( g'(x) = -1/(x-\pi/2)^2 \). Note that the limit will be taken as \( x \to (\pi/2)^- \).

Step 4: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule to compute \[ \ln L = \lim_{x \to (\pi/2)^-} \frac{-2 \sin(2x)}{\cos(2x)} \cdot \frac{(x-\pi/2)^2}{-1} \]. This simplifies to \( 2 \cdot \lim_{x \to (\pi/2)^-} (x-\pi/2) \tan(2x) \).

Step 5: Evaluate the Simplified Limit

As \( x \to (\pi/2)^- \), \( \tan(2x) \to 0 \) because \( \sin(2x) = 2(x-\pi/2) \), which makes the limit \( 2 \cdot 0 = 0 \). Therefore, \( \ln L = 0 \).

Step 6: Find the Limit of the Original Expression

Since \( \ln L = 0 \), taking the exponential of both sides gives \( L = e^0 = 1 \). Thus, the original limit \( \lim_{x \to (\pi/2)^-} (\cos(2x))^{x-\pi/2} = 1 \).

Key Concepts

Indeterminate Forms

Indeterminate forms in calculus describe expressions that lack a defined value or limit through standard algebraic manipulation. Such cases often arise during limit calculations and can seem perplexing at first. When you encounter expressions like

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0^0 \)
• \( \infty - \infty \)
• \( 0 \times \infty \)

these are prime examples of indeterminate forms. They require special techniques, such as l'Hôpital's Rule or algebraic manipulation, to evaluate their limits reliably.

In the problem with the limit \( \lim_{x \to (\pi/2)^-} (\cos 2x)^{x-\pi/2} \), we identified the form \( 0^0 \) as indeterminate. This means neither component approaches zero, leading to uncertainty about the overall behavior of the expression. Being aware of these indeterminate forms is crucial, as they signal that a typical limit evaluation may not work, nudging us toward applying techniques such as transforming via logarithms or using derivatives.

Limits in Calculus

Limits are fundamental to calculus, playing a crucial role in the definition of continuity, derivatives, and integrals. Understanding limits involves studying the behavior of a function as its input approaches a certain value. A limit expresses what value a function appears to be approaching when inputs near a particular point.

In examining limits, like in our exercise, we often deal with the left-hand limit \( x \to (\pi/2)^- \), indicating values of \( x \) approaching \( \pi/2 \) from the left. It's important to consider whether these limits exist or if they present indeterminate forms. Limits help us understand tendencies in calculus, tackling expressions that don't immediately yield answers by direct substitution.When indeterminate forms occur, l'Hôpital's Rule provides a powerful technique to evaluate the limit. This process involves derivates, helping us handle complex forms through simplified linear approximations that lead to clearer solutions.

Exponential and Logarithmic Differentiation

Exponential and logarithmic differentiation are techniques that help differentiate more complex functions. They are particularly handy when handling exponential functions and products or quotients posing as bases with variable exponents.

In the exercise, we applied logarithmic differentiation to tackle the form \( 0^0 \). By taking the natural logarithm of both sides, the leading-edge complexity reduces into a more manageable form. The equation \( \ln L = \lim_{x \to (\pi/2)^-} (x - \pi/2) \ln(\cos(2x)) \) transforms a power expression into a product that can subsequently be simplified using calculus techniques.Within this framework, differentiation becomes easier, leveraging properties of logarithms to break down the expressions further. Logarithmic rules such as

• The product rule \( \ln(a \cdot b) = \ln a + \ln b \)
• The power rule \( \ln(a^b) = b \ln a \)

become particularly beneficial in resolving the limits. By combining these strategies with l'Hôpital's Rule, the final evaluation resulted in a clarified understanding of the original expression's limit, showing its value equals 1.