Problem 92 - Thomas Calculus in SI Units

Question

Find the limits. $$\lim _{x \rightarrow 1^{+}} \frac{\sqrt{x^{2}-1}}{\sec ^{-1} x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Interpret the Limit Expression

We are tasked with finding the right-hand limit of the expression \( \frac{\sqrt{x^2 - 1}}{\sec^{-1}x} \) as \( x \to 1^+ \). This means we consider values of \( x \) approaching 1 from the right.

Step 2: Simplify the Expression by Analyzing Components

First, break down the components of the expression: \( \sqrt{x^2 - 1} \) and \( \sec^{-1}x \). Since we approach from the right, \( x^2 - 1 \) tends towards zero and is non-negative, and thus \( \sqrt{x^2 - 1} \to 0 \) as \( x \to 1^+ \). Also, \( \sec^{-1}x \) is defined and goes to zero.

Step 3: Behavior of Inverse Secant Function

As \( x \to 1^+ \), the inverse secant \( \sec^{-1}x \) approaches \( \sec^{-1}(1) = 0 \). So, the denominator approaches zero.

Step 4: Apply L'Hôpital's Rule

Since both the numerator \( \sqrt{x^2 - 1} \) and the denominator \( \sec^{-1}x \) approach zero as \( x \to 1^+ \), the limit is in the form \( \frac{0}{0} \). This allows us to apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately and then finding the limit.

Step 5: Differentiate the Components Separately

Differentiate the numerator: \( f(x) = \sqrt{x^2 - 1} \) gives \( f'(x) = \frac{x}{\sqrt{x^2 - 1}} \). Differentiate the denominator: \( g(x) = \sec^{-1}x \) gives \( g'(x) = \frac{1}{x\sqrt{x^2 - 1}} \) using the chain rule.

Step 6: Calculate the Limit Using L'Hôpital's Rule

By L'Hôpital's Rule, calculate:\[\lim _{x \rightarrow 1^{+}} \frac{f'(x)}{g'(x)} = \lim _{x \rightarrow 1^{+}} \frac{\frac{x}{\sqrt{x^2 - 1}}}{\frac{1}{x\sqrt{x^2 - 1}}} = \lim _{x \rightarrow 1^{+}} x^2 = 1.\]Thus, the limit is 1.

Key Concepts

L'Hôpital's Rule

When evaluating limits, sometimes direct substitution yields an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). L'Hôpital's Rule provides a way to tackle these challenges. It states that if \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \) or both approach infinity, then:

• \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this limit exists.

To apply the rule:

• Differentiation is performed on the numerator and the denominator separately.
• After differentiating, plug in the limit value to find the new limit.
• If the new limit remains indeterminate, repeat the process.

L'Hôpital's Rule does not solve limits that are not indeterminate forms, so ensure \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) is present before applying it.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the usual trigonometric functions. They are used to find angles when the trigonometric value is known. In our exercise, \( \sec^{-1} x \) is the inverse secant function.

• \( \sec^{-1} x \) is valid for \( x \geq 1 \) or \( x \leq -1 \).
• As \( x \to 1^+ \), \( \sec^{-1} x \) approaches 0, because \( \sec^{-1}(1) = 0 \).
• Inverse functions often require careful attention to their domain and range for solving limits accurately.

Understanding the behaviors of these functions is crucial, especially in limits where their output significantly impacts the final result. Recognizing how \( \sec^{-1} x \) behaves as \( x \) approaches certain critical points allows you to anticipate and solve limit problems effectively without confusion.

Differentiation Techniques

Differentiation is a fundamental tool in calculus used to find the rate at which a function changes at a point. When applying L'Hôpital's Rule, proper differentiation techniques come into play:

• The Chain Rule: Used when differentiating composite functions, like \( \sqrt{x^2 - 1} \), which results in \( \frac{x}{\sqrt{x^2 - 1}} \).
• Derivative of Inverse Functions: For \( \sec^{-1} x \), we use its known derivative formula: \( g'(x) = \frac{1}{x\sqrt{x^2-1}} \).
• Ensure the derivatives are correctly applied and simplified before finding their limit.

These techniques not only aid in computing derivatives but also help in simplifying and solving limit problems efficiently. Mastering these methods makes complex calculus problems more accessible.