Problem 59 - Calculus Single Variable

Question

Calculate the given limit. \(\lim _{x \rightarrow 0} \frac{\arcsin (x)-x}{\arctan (x)-x}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Indeterminate Form

Recognize that as \( x \to 0 \), both the numerator \( \arcsin(x) - x \) and the denominator \( \arctan(x) - x \) tend to \( 0 \). This confirms we have a \( \frac{0}{0} \) indeterminate form, which is suitable for applying L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) indeterminate form, we can use L'Hôpital's Rule. Let's differentiate the numerator and the denominator separately:- Derivative of the numerator: \( \frac{d}{dx}[\arcsin(x) - x] = \frac{1}{\sqrt{1-x^2}} - 1 \).- Derivative of the denominator: \( \frac{d}{dx}[\arctan(x) - x] = \frac{1}{1+x^2} - 1 \).

Step 3: Simplify the New Limit

Substitute the derivatives back into the limit:\[\lim_{x \to 0} \frac{\frac{1}{\sqrt{1-x^2}} - 1}{\frac{1}{1+x^2} - 1}\]This requires further simplification to be evaluated as \(x\) approaches 0.

Step 4: Evaluate the Limit as x Approaches 0

Evaluate the limit as \( x \to 0 \):1. In the numerator, \( \frac{1}{\sqrt{1-x^2}} - 1 \to 0 \) as \( x \to 0 \). This simplifies further using a Taylor series approximation, but at \( x = 0 \), it remains 0.2. In the denominator, \( \frac{1}{1+x^2} - 1 \to 0 \) as \( x \to 0 \), which also simplifies to 0.Evaluate each and find that the derivatives are proportional around \( x = 0 \), simplifying to \( \frac{1/2}{1/2} = 1 \).

Step 5: Conclusion

After evaluating, the limit is calculated as:\[\lim_{x \to 0} \frac{1}{1} = 1\]This indicates that the limit evaluates to 1.

Key Concepts

Indeterminate Forms

When evaluating limits, encountering an indeterminate form can initially seem puzzling. However, it often signals the potential to use more advanced techniques like L'Hôpital's Rule. Indeterminate forms occur in limits that don't explicitly lead to a numerical value. One common indeterminate form is \( \frac{0}{0} \), which means direct substitution in the limit does not give a clear result.

In the given exercise, as \( x \to 0 \), both the expressions \( \arcsin(x) - x \) and \( \arctan(x) - x \) tend to zero. This forms a \( \frac{0}{0} \) situation—a classic indeterminate form. Recognizing this helps you decide from the start that L'Hôpital's Rule is applicable.

Derivatives

Derivatives play a key role in applying L'Hôpital's Rule effectively. In the context of limits, we use derivatives to transform the expression into something more manageable.

• The derivative of the numerator \( \arcsin(x) - x \) is found by differentiating each part: \( \frac{d}{dx} [\arcsin(x) - x] = \frac{1}{\sqrt{1-x^2}} - 1 \).
• Similarly, the derivative of the denominator \( \arctan(x) - x \) is \( \frac{d}{dx} [\arctan(x) - x] = \frac{1}{1+x^2} - 1 \).

These derivatives allow us to rewrite the original limit expression, simplifying the complex fraction and making the limit evaluation at \( x = 0 \) clearer. Derivatives thus help transform a \( \frac{0}{0} \) indeterminate form into an expression that can be properly evaluated.

Trigonometric Limits

Trigonometric limits often involve functions like sine, cosine, tangent, and their inverse functions. These can create more complex challenge points when they appear in indeterminate forms. In solving trigonometric limits, understanding the behavior of inverse trigonometric functions is essential. Specifically:

• For small values of \( x \), \( \arcsin(x) \) can be approximated by \( x \). Therefore, \( \arcsin(x) - x \) approaches zero as \( x \to 0 \).
• Similarly, \( \arctan(x) \) is approximately \( x \) for small \( x \), leading \( \arctan(x) - x \) also to approach zero as \( x \to 0 \).

Using these approximations can sometimes simplify solving limits, but applying derivatives as per L'Hôpital's Rule is often a more systematic approach.

Limit Evaluation

Evaluating limits can be seen as the process of determining the behavior of a function as the input approaches a particular value. For the given problem, evaluating the limit after applying L'Hôpital's Rule involves checking the behavior of the function created by the derivatives.

After differentiating, you obtain a new limit:\[\lim_{x \to 0} \frac{\frac{1}{\sqrt{1-x^2}} - 1}{\frac{1}{1+x^2} - 1}\]Both the numerator and the denominator simplify to zero as \( x \to 0 \). Continuing with derivative approximations or power series expansion helps show proportional simplification resulting in further simplification to unity: \( \frac{1/2}{1/2} = 1 \).

Thus, the limit evaluates to 1, demonstrating a seamlessly managed transition from a complex expression to a simple answer. This showcases the power of calculus tools in simplifying and solving intricate problems.