Problem 99 - Thomas Calculus

Question

Find the limits in Exercises \(97-104\) $$ \lim _{x \rightarrow \infty} x \tan ^{-1} \frac{2}{x} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify the Limit Problem

The problem requires us to evaluate the limit \( \lim_{x \rightarrow \infty} x \tan^{-1} \frac{2}{x} \). This indicates that we want to find the behavior of the function \( x \tan^{-1} \frac{2}{x} \) as \( x \) approaches infinity.

Step 2: Simplify the Expression Using Substitution

We substitute \( y = \frac{2}{x} \), which implies \( x = \frac{2}{y} \). As \( x \to \infty \), \( y \to 0^+ \). This simplifies the expression to \( \lim_{y \rightarrow 0^+} \frac{2}{y} \tan^{-1}(y) \).

Step 3: Use Taylor Series Expansion

The Taylor series expansion for \( \tan^{-1}(y) \) around \( y = 0 \) is \( y - \frac{y^3}{3} + \cdots \) . For very small \( y \), \( \tan^{-1}(y) \approx y \). Thus, our expression becomes \( \frac{2}{y} \cdot y = 2 \).

Step 4: Calculate the Limit

Using the approximation \( \tan^{-1}(y) \approx y \), substitute back to find \( \lim_{y \to 0^+} \frac{2}{y} \times y = \lim_{y \to 0^+} 2 = 2 \).

Step 5: Conclusion

After performing the limit calculation using substitution and approximation, the final result of the limit \( \lim_{x \rightarrow \infty} x \tan^{-1} \frac{2}{x} \) is 2.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the functions that reverse the effect of the standard trigonometric functions like sine, cosine, and tangent. They are crucial when you need to determine the angle given a trigonometric ratio.

In our exercise, we deal with the inverse tangent function, denoted as \( \tan^{-1}(x) \). This function provides the angle whose tangent is \( x \). It's handy for solving equations where the tangent of an angle is known, but the angle is not.

• The inverse tangent function, \( \tan^{-1}(x) \), has a range of \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
• It's an odd function, meaning \( \tan^{-1}(-x) = -\tan^{-1}(x) \).
• Near zero, \( \tan^{-1}(x) \approx x \), which is a simplification useful for Taylor series expansion.

Taylor Series

The Taylor series is a mathematical representation that allows functions to be expressed as infinite sums of terms. Each term is based on the function's derivatives evaluated at a specific point. This makes it invaluable when approximating functions near that point.

In the given exercise, the Taylor series expansion of \( \tan^{-1}(y) \) is particularly useful as \( y \) approaches zero. Using the expansion:

• \( \tan^{-1}(y) \approx y \) for small values of \( y \). This is because higher-order terms like \( \frac{y^3}{3} \) become negligible.
• The approximation massively simplifies calculations, making it easier to solve for limits.

This is why, in the solution, the expression \( x \tan^{-1} \frac{2}{x} \) simplifies beautifully as \( x \to \infty \). By substituting \( y = \frac{2}{x} \), we can simplify the limit expression to something much more digestible.

Infinity Limits

Infinity limits help us understand how functions behave as inputs grow unbounded in size. Mathematically, limits approaching infinity explore the asymptotic behavior of functions.

In our limit problem, as \( x \to \infty \), the transformation to \( y = \frac{2}{x} \) allows us to reframe the problem. Instead of evaluating trigonometric functions at increasingly large numbers, we reimagine it at values shrinking towards zero.

• The use of a simple substitution transforms what might first appear as a complex problem into something more tractable.
• As \( y \to 0^+ \), the Taylor approximation shows that \( \tan^{-1}(y) \approx y \), maintaining simplicity even towards these boundary values.

Infinity limits, coupled with smart substitutions and series expansions, offer powerful techniques to solve otherwise challenging problems effectively.