Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{\tan ^{-1} 3 x}{\sin ^{-1} x} $$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: Understanding Inverse Functions

We begin by looking at the functions involved: \( \tan^{-1}(3x) \) is the inverse tangent function and \( \sin^{-1}(x) \) is the inverse sine function. Both functions tend toward zero as \( x \) approaches zero.

Step 2: Applying L'Hôpital's Rule

Since this limit is in the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states that for functions \( f(x) \) and \( g(x) \), if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right side exists.

Step 3: Differentiate Numerator and Denominator

Differentiate \( \tan^{-1}(3x) \) with respect to \( x \): \( \frac{d}{dx} [\tan^{-1}(3x)] = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2} \). Differentiate \( \sin^{-1}(x) \) with respect to \( x \): \( \frac{d}{dx} [\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}} \).

Step 4: Evaluate the New Limit

Using L'Hôpital's Rule, evaluate the limit: \( \lim_{x \to 0} \frac{\frac{3}{1 + 9x^2}}{\frac{1}{\sqrt{1 - x^2}}} = \lim_{x \to 0} \frac{3 \sqrt{1 - x^2}}{1 + 9x^2} \). As \( x \to 0 \), this simplifies to \( \frac{3 \cdot 1}{1 + 0} = 3 \).

Step 5: Conclusion of the Problem

The limit has been evaluated correctly step-by-step using L'Hôpital's Rule and simplifying the expression as \( x \to 0 \). The final result of the limit is 3.

Key Concepts

Inverse trigonometric functions

Inverse trigonometric functions, such as \( \tan^{-1}(x) \) and \( \sin^{-1}(x) \), are essential in mathematics because they provide the angle that corresponds to a given trigonometric ratio. These functions are primarily used to determine the angles when the values of sine or tangent are known.

Both \( \tan^{-1} \) and \( \sin^{-1} \) have specific ranges:

• \( \tan^{-1}(x) \) gives an angle in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
• \( \sin^{-1}(x) \) produces an angle within the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

As \( x \) approaches 0, both functions also tend toward zero. This tendency is vital when analyzing limits in calculus, as seen in the given example. The subtle behaviors of inverse trigonometric functions at boundaries or specific points often lead to indeterminate forms, requiring special rules or techniques to resolve.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique in calculus for resolving limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Named after the French mathematician Guillaume de l'Hôpital, this rule simplifies the process of calculating limits by relating them to the derivatives of their numerators and denominators.

When faced with an indeterminate form, you can apply this rule by:

• Taking the derivative of the numerator \( f(x) \).
• Taking the derivative of the denominator \( g(x) \).
• Evaluating the limit of \( \frac{f'(x)}{g'(x)} \) as \( x \) approaches the target value.

This approach is only valid if the limit of the derivatives exists or results in a determinate form. In the specific exercise, applying L'Hôpital's Rule effectively transformed the problem from tackling the raw \( \frac{0}{0} \) form to a more manageable expression that could be evaluated.

It is an indispensable tool when dealing with complex functions and is especially helpful when dealing with inverse trigonometric functions, which often induce these types of indeterminate forms.

Differentiation

Differentiation is the mathematical process of determining the rate at which a function is changing at any given point. It involves finding the derivative of a function. In our example, differentiation is essential for applying L'Hôpital's Rule and getting the desired limit solved.

Derivatives have specific formulas for different functions:

• The derivative of \( \tan^{-1}(3x) \) is found using the chain rule, yielding \( \frac{3}{1 + (3x)^2} \), which simplifies to \( \frac{3}{1 + 9x^2} \).
• For \( \sin^{-1}(x) \), the derivative is \( \frac{1}{\sqrt{1-x^2}} \).

These derivatives tell us how the inverse trigonometric functions behave near zero, which is crucial for applying L'Hôpital's Rule. Differentiation simplifies the process by reducing the complexity of the original expression, allowing us to evaluate challenging limits by replacing the original indeterminate form with a determinate one.