Problem 19 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x}{x}$$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Evaluate Direct Substitution

To find \( \lim_{x \rightarrow 0} \frac{\sin^{-1}(2x)}{x} \), we start by directly substituting \(x = 0\) into the expression. We get \( \frac{\sin^{-1}(0)}{0} = \frac{0}{0} \). This is an indeterminate form, which means we need another method to evaluate the limit.

Step 2: Apply L'Hôpital's Rule

Since direct substitution resulted in an indeterminate form \( \frac{0}{0} \), L'Hôpital's Rule is applicable. L'Hôpital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \pm \infty / \pm \infty \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) provided the limit on the right exists.

Step 3: Differentiate the Numerator and Denominator

To apply L'Hôpital's Rule, we need to differentiate the numerator \( \sin^{-1}(2x) \) and the denominator \(x\). The derivative of \( \sin^{-1}(u) \) with respect to \(u\) is \( \frac{1}{\sqrt{1-u^2}} \). Therefore, the derivative of \( \sin^{-1}(2x) \) with respect to \(x\) is \[ f'(x) = \frac{2}{\sqrt{1 - (2x)^2}} \].The derivative of \(x\) is \(g'(x)=1\).

Step 4: Evaluate the Limit of Derivatives

Substitute the derivatives back into the limit:\[ \lim_{x \to 0} \frac{2}{\sqrt{1 - (2x)^2}} \].Evaluate this limit by substituting \(x = 0\):\[ \frac{2}{\sqrt{1 - 0^2}} = 2 \].

Step 5: Conclude the Result

The value of the limit, after applying L'Hôpital's Rule and differentiating, is 2. Therefore, the limit \( \lim_{x \rightarrow 0} \frac{\sin^{-1}(2x)}{x} \) is 2.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule helps simplify complex expressions and makes solving limits possible when direct substitution doesn't work.
The rule states that if you have a limit of the form \( \lim_{x \to a} \frac{f(x)}{g(x)} \) that results in an indeterminate form, you can take the derivatives of the numerator and the denominator separately. Then, evaluate the new limit \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided this limit exists.
In our problem, evaluating \( \lim_{x \rightarrow 0} \frac{\sin^{-1}(2x)}{x} \) gives \( \frac{0}{0} \), an indeterminate form. By applying L'Hôpital's Rule, we differentiate both parts: the numerator \( \sin^{-1}(2x) \) becomes \( \frac{2}{\sqrt{1 - (2x)^2}} \), while the denominator \( x \) becomes 1. The rule then guides us to find that the limit equals 2.

Indeterminate Forms

Indeterminate forms occur in limit problems when direct substitution leads to perplexing results like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms signal that the limit requires a different approach, often through algebraic manipulation or using calculus rules like L'Hôpital's.
When a student encounters an indeterminate form, it means that while direct evaluation at the point of interest does not yield a number, there might still be a limit if analyzed differently. In the example exercise, substituting \( x = 0 \) into \( \frac{\sin^{-1}(2x)}{x} \) results in \( \frac{0}{0} \), indicating an indeterminate form.
This form prompts the application of L'Hôpital's Rule to resolve the limit by differentiating both the numerator and the denominator, allowing us to discover the true behavior of the function as \( x \) approaches the limit point.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the roles of the sine, cosine, and tangent functions. For example, \( \sin^{-1}(x) \) or arcsin returns the angle whose sine is \( x \). These functions are commonly used in calculus to solve problems involving angles and arc lengths.
A key aspect to remember when dealing with these functions is their derivatives. For \( \sin^{-1}(u) \), the derivative with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \). This formula is vital when applying calculus techniques like differentiation to evaluate limits or analyze function behavior near specific points.
In the exercise, understanding the derivative of \( \sin^{-1}(2x) \) becomes crucial for using L'Hôpital's Rule. The derivative is \( \frac{2}{\sqrt{1 - (2x)^2}} \), which simplifies resolving the limit and achieving the final answer.