Question

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

Short answer

The limit is 0.

Step-by-step solution

Substitute the Limit Value into the Expression

Substitute \( x = \frac{\pi}{2} \) into \( \tan(x) - \sec(x) \) to check for an indeterminate form: \( \tan\left(\frac{\pi}{2}\right) - \sec\left(\frac{\pi}{2}\right) \).

Recognize the Indeterminate Form

As \( \tan(\frac{\pi}{2}) = \infty \) and \( \sec(\frac{\pi}{2}) = \infty \), the expression \( \tan(x) - \sec(x) \) approaches \( \infty - \infty \), which is an indeterminate form.

Rewrite the Expression Using Trigonometric Identities

Rewrite \( \tan(x) - \sec(x) \) as \( \frac{\sin(x)}{\cos(x)} - \frac{1}{\cos(x)} \).

Combine into a Single Fraction

Combine into a single fraction: \( \frac{\sin(x) - 1}{\cos(x)} \).

Apply l'Hôpital's Rule

Since the expression is in the indeterminate form \( \frac{0}{0} \) as \( x \rightarrow \frac{\pi}{2} \), apply l'Hôpital's Rule by differentiating the numerator and the denominator. The derivative of \( \sin(x) - 1 \) is \( \cos(x) \) and the derivative of \( \cos(x) \) is \(-\sin(x)\).

Reevaluate the Limit Using the Derivatives

Re-evaluate the limit: \( \lim_{x \rightarrow \pi/2} \frac{\cos(x)}{-\sin(x)} \). As \( x \to \frac{\pi}{2} \), \( \cos(x) \rightarrow 0 \) and \( \sin(x) \rightarrow 1 \).

Calculate the Limit

Substitute the limit values into the new expression: \( \frac{\cos(\frac{\pi}{2})}{-\sin(\frac{\pi}{2})} = \frac{0}{-1} = 0 \). The limit is therefore 0.

Key concepts

Indeterminate Forms

In calculus, you often encounter situations where directly evaluating a limit gives you a result like \( \frac{0}{0} \) or \( \infty - \infty \). These are called indeterminate forms. They occur because substitution leads to an undefined or ambiguous situation.

The exercise involves evaluating \( \lim_{x \to \pi/2} (\tan(x) - \sec(x)) \). When you plug \( x = \pi/2 \) into this expression, you get \( \tan(\pi/2) - \sec(\pi/2) \), which evaluates to \( \infty - \infty \). This is an indeterminate form because it suggests an ambiguous infinite result.

• Common Indeterminate Forms: \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), \( 0 \cdot \infty \), \( 1^\infty \), \( 0^0 \), \( \infty^0 \).
• Importance: Recognizing indeterminate forms is essential for applying techniques like l'Hôpital's Rule, which can help resolve these into meaningful numeric limits.

Trigonometric Limits

When you're working with trigonometric functions like \( \tan(x) \) and \( \sec(x) \), limits are used to analyze behavior as the variable approaches a certain point. Trigonometric limits are a core part of calculus, dealing specifically with functions involving angles.

In this exercise, rewriting the expression \( \tan(x) - \sec(x) \) as \( \frac{\sin(x)}{\cos(x)} - \frac{1}{\cos(x)} \) helps find a common form. This manipulation uses basic trig identities, making the expression simpler for limit evaluation.

• Key Trigonometric Identities: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), \( \sec(x) = \frac{1}{\cos(x)} \).
• Simplifying Limits: Combining expressions into a single fraction often makes it easier to assess the limit's behavior.
• Relevance: Familiarity with trigonometric identities is crucial in simplifying complex expressions and applying limit-solving techniques.

Calculus Problems

Calculus allows us to solve problems by analyzing changes and behaviors of functions, often using limits. It's the foundation for l'Hôpital's Rule, which helps resolve indeterminate forms when calculating limits.

In this problem, the initial expression \( \frac{\sin(x) - 1}{\cos(x)} \) yields an indeterminate form \( \frac{0}{0} \) when \( x \to \pi/2 \). Applying l'Hôpital's Rule means differentiating the numerator and the denominator. This step transforms the limit into \( \lim_{x \to \pi/2} \frac{\cos(x)}{-\sin(x)} \), making it solvable.

• L'Hôpital's Rule: If a limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the top and bottom, then re-evaluate the limit.
• Why It Works: Differentiation helps bypass the undefined form, offering a new expression that often leads to an evaluable limit.
• Calculation Outcome: Substituting the limit gives \( \frac{0}{-1} = 0 \), showing that the original indeterminate form resulted in a final limit of 0.