Problem 95 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Express the Limit

We need to find \( \lim _{x \rightarrow 0} \frac{\tan^{-1} x^{2}}{x \sin^{-1} x} \). This is originally in indeterminate form \(\frac{0}{0}\), so let's use L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately if the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). So, we need to compute the derivatives:- Derivative of \(\tan^{-1} x^2\) is \(\frac{d}{dx} \tan^{-1} x^2 = \frac{2x}{1+(x^2)^2}\).- Derivative of \(x \sin^{-1} x\) is \( \frac{d}{dx} (x \sin^{-1} x) = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \).

Step 3: Simplify and Re-evaluate the Limit

The new limit is: \[\lim _{x \rightarrow 0} \frac{\frac{2x}{1+x^4}}{\sin^{-1} x + \frac{x}{\sqrt{1-x^2}}}\]Evaluate this at \(x = 0\). The expression simplifies because \( \sin^{-1} x \approx x\) and \( \frac{x}{\sqrt{1-x^2}} \approx x\) as \( x \rightarrow 0\). Hence, the denominator becomes \(2x\).

Step 4: Calculate the Limit

Simplify the limit: \[\lim_{x \to 0} \frac{\frac{2x}{1+x^4}}{2x} = \lim_{x \to 0} \frac{2x}{2x(1+x^4)}\]This reduces to \(\lim_{x \to 0} \frac{1}{1+x^4} = 1 \) since the \(x^4\) term vanishes as \(x\) approaches zero.

Key Concepts

Limit Evaluation

Finding limits is a crucial topic in calculus. A limit expression, such as \[ \lim_{x \to 0} \frac{\tan^{-1} x^2}{x \sin^{-1} x} \]helps determine the value a function approaches as the variable gets extremely close to a certain point.

In this case, we investigate behavior near zero, providing insights into function behavior. Key steps for evaluating limits often involve:

• Simplifying the expression
• Substituting values (when they make sense)
• Using L'Hôpital's Rule if necessary

Through these methods, you'll often resolve apparent undecidable forms, providing clarity in otherwise complicated expressions.

Indeterminate Forms

Indeterminate forms arise when directly substituting a value into a limit expression doesn't immediately provide a meaningful result. They often take forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

In the exercise, upon inserting \(x = 0\) into \( \frac{\tan^{-1} x^2}{x \sin^{-1} x} \), you encounter \(\frac{0}{0}\). This is a classic example of an indeterminate form.

When faced with such forms, L'Hôpital's Rule is your ally. It guides you to differentiate the numerator and denominator separately. Doing so can transform an indeterminate form into one where direct substitution provides meaningful results. Hence, understanding indeterminate forms is crucial for progressing in calculus problem-solving.

Calculus Problem Solving

Solving calculus problems like evaluating limits often requires combining a toolkit of approaches. These typically include:

• Understanding the behavior of functions near certain points
• Simplifying expressions
• Using calculus rules, such as L'Hôpital's Rule

The strategy in the provided problem involves recognizing the potential use of calculus techniques.

By pinpointing when to apply differentiation or algebraic adjustments, your calculus problem-solving skills become refined. This not only aids in understanding complex problems but also prepares you for tackling a range of similar challenges in mathematics. Each problem resolved enhances intuition and sharpens reasoning abilities, setting a strong foundation for further mathematical exploration.