Problem 93 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Recognize the Indeterminate Form

First, identify that the expression \(x \tan^{-1} \frac{2}{x}\) has a potential indeterminate form. As \(x\) approaches infinity, \(\frac{2}{x}\) approaches 0, making \(\tan^{-1} \frac{2}{x}\) approach \(\tan^{-1}(0)\), which is 0. This gives an indeterminate form of \(\infty \times 0\).

Step 2: Rewrite the Expression as a Quotient

Rewrite the expression \(x \tan^{-1} \left(\frac{2}{x}\right)\) in the form of a quotient: \(\frac{\tan^{-1}\left(\frac{2}{x}\right)}{\frac{1}{x}}\). This helps us use L'Hôpital's Rule since both numerator and denominator approach 0 as \(x\) approaches infinity.

Step 3: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to \(\lim_{x \to \infty} \frac{\tan^{-1}\left(\frac{2}{x}\right)}{\frac{1}{x}}\). Both the numerator and denominator approach 0, so we can differentiate the numerator and the denominator.

Step 4: Differentiate the Numerator and Denominator

Differentiate the numerator \(\tan^{-1}\left(\frac{2}{x}\right)\) using the chain rule. The derivative is \(-\frac{2}{x^2 + 4}\). Differentiate the denominator \(\frac{1}{x}\), which gives \(-\frac{1}{x^2}\).

Step 5: Take the Limit of the New Expression

After applying L'Hôpital's Rule, the expression becomes \(\lim_{x \to \infty} \frac{-\frac{2}{x^2 + 4}}{-\frac{1}{x^2}}\). Simplify this to \(\lim_{x \to \infty} \frac{2x^2}{x^2 + 4}\).

Step 6: Evaluate the Final Limit

As \(x\) approaches infinity, the dominant term in the numerator and the denominator is \(x^2\). Therefore, the expression \(\frac{2x^2}{x^2 + 4}\) simplifies to \(\frac{2}{1} = 2\). Hence, the limit is 2.

Key Concepts

L'Hôpital's Rule

Suppose you're faced with a limit problem where both the numerator and the denominator of a fraction tend to zero or infinity at the same time. In such a case, L'Hôpital's Rule is your tool to resolve the limit. Named after the French mathematician Guillaume de l'Hôpital, this rule states that you can differentiate the numerator and the denominator separately and then take the limit again. However, keep in mind:

• L'Hôpital's Rule can only be applied to indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• You'll differentiate the top and the bottom of the fraction and check if the new limit is easier to compute.
• Repeat the process if necessary until a determinate form appears.

When you employ this rule effectively, complex limits can often be simplified to easily recognizable forms.

Indeterminate Forms

Indeterminate forms occur when evaluating limits leads to expressions that don't have an obvious value. In our example, the expression initially evaluated to \(\infty \times 0\). This is one of several indeterminate forms. Here are some common indeterminate forms you might come across in calculus:

• \(0/0\)
• \(\infty/\infty\)
• \(0 \times \infty\)
• \(\infty - \infty\)
• \(1^\infty\)
• \(0^0\)
• \(\infty^0\)

Indeterminate forms signal the need for further analysis, often involving L'Hôpital's Rule or algebraic manipulation, to evaluate a limit. Recognizing these forms is critical to determine the correct method to find solutions.

Arctangent Function

The arctangent function, often symbolized as \(\tan^{-1}x\), is the inverse function of the tangent. This trigonometric function helps find angles whose tangent is known. Here are some key aspects to consider:

• The range of \(\tan^{-1}x\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• \(\tan^{-1}(0) = 0\) is a fact used in calculations.

The arctangent function's behavior is crucial to understanding limits, especially when \(\tan^{-1}\left(\frac{2}{x}\right)\) is involved, due to \(\frac{2}{x}\) approaching zero as \(x\) approaches infinity.

Chain Rule

The chain rule is a fundamental tool in differentiation, especially when dealing with composite functions. It provides a way to differentiate functions nested within other functions, which we saw in our example.

When differentiating \(\tan^{-1}\left(\frac{2}{x}\right)\), we used the chain rule by:

• Identifying the outer function as \(\tan^{-1}(u)\) and the inner function as \(u = \frac{2}{x}\).
• Finding the derivative of the outer function: \(\frac{d}{du} \tan^{-1}u = \frac{1}{1+u^2}\).
• Finding the derivative of the inner function \(\frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}\).
• Applying the chain rule: the derivative becomes \(-\frac{2}{x^2 + 4}\).

Mastering the chain rule is essential, as it appears frequently and simplifies the process of differentiation in many calculus problems.