Problem 54 - Calculus for Biology and Medicine

Question

Find the limits in Problems Be sure to check whether you can apply I'Hospital's rule before you evaluate the limit. $$ \lim _{x \rightarrow(\pi / 2)^{-}} \frac{\tan x}{1+\sec x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Form

We start by substituting \( x = \frac{\pi}{2}^{-} \) into the expression \( \frac{\tan x}{1 + \sec x} \). We find that as \( x \) approaches \( \frac{\pi}{2}^{-} \), \( \tan x \to \infty \) and \( \sec x \to \infty \) because the cosine function approaches zero from the positive side. Therefore, the form is \( \frac{\infty}{\infty} \), which allows us to use L'Hospital's Rule.

Step 2: Apply L'Hospital's Rule

According to L'Hospital's Rule, we can differentiate the numerator and the denominator until we resolve the indeterminate form. The derivative of \( \tan x \) is \( \sec^2 x \), and the derivative of \( 1 + \sec x \) is \( \sec x \tan x \). Thus, the limit becomes:\[\lim _{x \rightarrow(\pi / 2)^{-}} \frac{\sec^2 x}{\sec x \tan x}\]

Step 3: Simplify the Expression

We can simplify \( \frac{\sec^2 x}{\sec x \tan x} \) to \( \frac{\sec x}{\tan x} \) or equivalently \( \frac{\sec x}{\frac{\sin x}{\cos x}} = \frac{\sec x \cos x}{\sin x} \). Since \( \sec x = \frac{1}{\cos x} \), this reduces to \( \frac{1}{\sin x} \).

Step 4: Evaluate the Limit

Substitute \( x = \frac{\pi}{2}^{-} \) into \( \frac{1}{\sin x} \). As \( x \to \frac{\pi}{2}^{-} \), \( \sin x \to 1 \). Therefore, the limit becomes:\[\lim _{x \rightarrow(\pi / 2)^{-}} \frac{1}{\sin x} = 1\]

Step 5: Conclusion

After applying L'Hospital's Rule and simplifying the expression, we evaluated the limit and found that it equals 1.

Key Concepts

Limit Evaluation

Limit evaluation is an important concept in calculus, helping us to understand the behavior of functions as they approach specific points. When we talk about evaluating limits, it involves analyzing what happens to a function's value as the input gets closer to a certain point, often referred to as the limit point. To evaluate limits, different techniques can be used:

• Direct Substitution: Plugging the value into the function.
• Factorization: Factoring expressions to simplify.
• Rationalization: Eliminating radicals or complex fractions.
• L'Hospital's Rule: Used for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

In this given problem, we used L'Hospital's Rule after identifying an indeterminate form. This helped in transforming the complex expression into a simpler form. Always check if the direct substitution leads to an undefined expression, which requires further techniques. Mastery of various limit evaluation techniques broadens your ability to solve calculus problems effectively.

Trigonometric Functions

Trigonometric functions are pivotal in evaluating limits, especially when dealing with periodic behaviors or angles. Common trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), tangent (\( \tan \)), and secant (\( \sec \)) among others. These functions behave in specific ways as they approach notable angles like 0, \( \pi/2 \), \( \pi \), etc.

• For \( \tan x \), as \( x \) approaches \( \pi/2 \) from the left, it trends toward infinity.
• For \( \sec x \), since it's the reciprocal of \( \cos x \), it also trends toward infinity when \( x \to \pi/2 \) from the negative side as \( \cos x \to 0 \).

Understanding these behaviors is crucial in problems involving limits of trigonometric functions. They enable us to predict and compute values as a function nears a specific point. Use trigonometric identities to simplify expressions when needed, aiding in comprehending how these functions react under limit conditions.

Indeterminate Forms

Indeterminate forms occur in limit problems where direct substitution doesn’t result in a clear or finite value. These forms often appear in the guise of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) among others.When facing such a problem, special rules like L'Hospital's Rule become invaluable. L'Hospital's Rule allows you to differentiate the numerator and denominator separately wherever these indeterminate forms appear, rendering a more manageable limit evaluation. Consider the original problem: substituting directly results in \( \frac{\tan(\pi/2^-)}{1+\sec(\pi/2^-)} = \frac{\infty}{\infty} \), an indeterminate form. By applying L'Hospital's Rule, you transform this complex form, providing a clear path to resolving the limit. Recognizing these forms is a skill developed through practice and aids greatly in progressing through calculus problems efficiently.