Problem 28 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow \infty} x \tan (1 / x)$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Evaluate the Form of the Limit

As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Therefore, the expression inside the limit becomes \( x \tan\left(\frac{1}{x}\right) \to \infty \cdot \tan(0) \), which is \( \infty \cdot 0 \). This is an indeterminate form, which allows us to investigate further for potential simplifications or apply l'Hôpital's Rule if convenient.

Step 2: Approximate the Function Using Taylor Expansion

For small values of \( \theta \), \( \tan(\theta) \approx \theta \). Since \( \frac{1}{x} \) is small as \( x \to \infty \), we can approximate \( \tan\left(\frac{1}{x}\right) \approx \frac{1}{x} \). Hence, the expression \( x \tan\left(\frac{1}{x}\right) \approx x \cdot \frac{1}{x} = 1 \).

Step 3: Assess the Limit

From the approximation, we find that the limit simplifies to a constant value as \( x \to \infty \). Therefore, \( \lim_{x \to \infty} x \tan\left(\frac{1}{x}\right) = 1 \).

Step 4: Confirm with l'Hôpital's Rule (if necessary)

Since an elementary method provided a clear solution, applying l'Hôpital's Rule is redundant here. However, if we reconsider \( f(x) = x \tan\left(\frac{1}{x}\right) \), rewriting it as \( \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}} \), both numerator and denominator go to 0. Therefore, l'Hôpital's Rule could be applied by differentiating numerator and denominator.

Key Concepts

l'Hôpital's Rule

When we face limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), l'Hôpital's Rule can be a useful tool. This rule states that if the limits of both the numerator and denominator of a fraction approach zero or infinity, then the limit of the fraction can be found by differentiating the numerator and the denominator separately, and then taking the limit of this new fraction.To apply l'Hôpital's Rule effectively:

• Ensure the original limit problem results in an indeterminate form.
• Differentiate the numerator and denominator.
• Evaluate the new limit.

In our exercise, after some transformation, the original expression \( x \tan\left(\frac{1}{x}\right) \) can be viewed as a fraction \( \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}} \). As both the numerator and the denominator approach zero when \( x \to \infty \), l'Hôpital's Rule could technically be used, even though the exercise was simplified through other means.

Indeterminate Forms

Indeterminate forms occur when evaluating a limit results in an undefined expression. Common forms include \( \infty - \infty \), \( 0 \times \infty \), and \( \frac{0}{0} \). These scenarios don't directly give us information about the limit's value, requiring further analysis.Let's focus on our exercise, where the form \( \infty \cdot 0 \) arises. Initially, it might seem to contradict itself, as multiplying any number by zero results in zero, yet infinity by zero is not straightforward. Herein lies the value of calling it 'indeterminate,' meaning further work is required to find the true nature of the limit. By using algebraic manipulation or mathematical tools like l'Hôpital's Rule, we transform these expressions into something more recognizable to determine the limit’s value.

Taylor Expansion

Taylor expansion provides an excellent method for approximating functions and simplifying complex forms, especially when dealing with limits in calculus. It provides an expansion of a function into an infinite sum of terms calculated from the values of its derivatives at a single point.For instance, the Taylor series approximation for \( \tan(\theta) \) can be written as:\[ \tan(\theta) \approx \theta + \frac{\theta^3}{3} + \cdots \]When \( \theta \) is very small, this reduces to \( \tan(\theta) \approx \theta \). In our problem, since \( \frac{1}{x} \) becomes very small as \( x \to \infty \), we substitute \( \tan\left(\frac{1}{x}\right) \approx \frac{1}{x} \). Thus, simplifying the whole expression \( x \tan\left(\frac{1}{x}\right) \) to \( x \cdot \frac{1}{x} = 1 \). This power of Taylor expansion lies in making complex limits accessible through simple approximations.