Problem 40 - Calculus: Early Transcendentals

Question

Prove that \( \displaystyle \lim_{x \to 0^+}\sqrt{x} e^{\sin (\pi/x)} = 0 \).

Step-by-Step Solution

Answer

The limit is 0 by the Squeeze Theorem.

Step 1: Understanding the Limit Expression

Here, we need to evaluate \( \lim_{x \to 0^+} \sqrt{x} \cdot e^{\sin (\pi/x)} \). As \( x \to 0^+ \), \( \sqrt{x} \to 0 \), but we must consider how the expression \( e^{\sin (\pi/x)} \) behaves since \( \sin (\pi/x) \) oscillates between -1 and 1. The challenge is understanding the interplay between these two parts of the expression.

Step 2: Simplifying the Exponential Term

Since \( -1 \leq \sin(\pi/x) \leq 1 \) for all \( x eq 0 \), it follows that \( e^{-1} \leq e^{\sin(\pi/x)} \leq e^1 \). This establishes a bounded behavior of the exponential term near \( x \to 0^+ \). Thus, \( e^{\sin(\pi/x)} \) is bounded by these fixed values.

Step 3: Evaluating the Behavior of \(\sqrt{x}\)

As \( x \to 0^+ \), \( \sqrt{x} \to 0 \). The crucial aspect is that as \( \sqrt{x} \to 0 \), the product \( \sqrt{x} \cdot e^{\sin (\pi/x)} \) will be influenced significantly more by the \( \sqrt{x} \to 0 \) term than the bounded oscillations of \( e^{\sin(\pi/x)} \).

Step 4: Applying the Squeeze Theorem

Since we have \( e^{-1} \leq e^{\sin(\pi/x)} \leq e^1 \) and \( \sqrt{x} \to 0 \), consider the inequality: \( e^{-1} \cdot \sqrt{x} \leq \sqrt{x} \cdot e^{\sin (\pi/x)} \leq e^1 \cdot \sqrt{x} \). Both bounding expressions \( e^{-1} \cdot \sqrt{x} \to 0 \) and \( e^1 \cdot \sqrt{x} \to 0 \) as \( x \to 0^+ \), allowing us to apply the Squeeze Theorem: the limits of the bounds and the original function all lead to 0.

Step 5: Conclusion of the Limit

By using the Squeeze Theorem with the bounds on \( e^{\sin(\pi/x)} \) and the limit behavior of \( \sqrt{x} \), we conclude that \( \lim_{x \to 0^+} \sqrt{x} e^{\sin (\pi/x)} = 0 \). This is because the limiting behavior of \( \sqrt{x} \) to zero dominates the bounded oscillations of the exponential term.

Key Concepts

Limits

In calculus, limits are fundamental to understanding functions' behavior as they approach specific values. The exercise involves evaluating the limit \( \lim_{x \to 0^+} \sqrt{x} e^{\sin (\pi/x)} \).Here's the idea: as \( x \) approaches zero from the positive side (denoted "+"), we closely observe how the function behaves. The expression \( \sqrt{x} \) straightforwardly approaches 0 as \( x \) approaches 0 from the right.However, the second part of the expression, \( e^{\sin(\pi/x)} \), involves an oscillating term, complicating the analysis. By understanding limits, we can deduce that if we manage to "trap" or "squeeze" our function between two other functions that both approach a particular limit, then our function must approach that limit too. This principle is what the Squeeze Theorem utilizes, allowing us to conclude that the limit is indeed 0.

Exponential Function

The exponential function is known for its rapid growth or decay, expressed typically as \( e^x \). In our problem, the function transitions to \( e^{\sin (\pi/x)} \), which complicates the problem due to the sine function's oscillation.Key things to remember about exponential functions:

• They are continuous and smooth.
• They grow faster than polynomial functions as their argument becomes large.
• They are never zero for any finite x, staying positive at all times.

For \( e^{\sin(\pi/x)} \), it's crucial to note its behavior is determined by \( \sin(\pi/x) \), which fluctuates between -1 and 1. This results in \( e^{-1} \leq e^{\sin(\pi/x)} \leq e^1 \), showcasing its bounded nature. This bounding is essential since it prevents the exponential term from becoming too large, maintaining control in the evaluation of the limit.

Oscillating Functions

Oscillating functions repeatedly vary between maximum and minimum values over intervals. The sine function is a classical oscillating function, and in our exercise, \( \sin(\pi/x) \) is an oscillating term that's pivotal in finding the limit.Characteristics of oscillating functions:

• They lack a single direction of growth.
• They can determine limits' complexity due to their fluctuation.
• Bounded values ensure they don't escape certain boundaries.

In \( \sin(\pi/x) \), the frequency of oscillations becomes infinitely rapid as \( x \to 0^+ \). However, despite these rapid oscillations, the bounded nature of \( \sin(\pi/x) \) ensures that the expression \( e^{\sin(\pi/x)} \) remains within a predictable range, allowing analysis through the Squeeze Theorem.

Bounded Functions

A bounded function is one where the outputs remain within a certain fixed range, which is critical when dealing with limits.What you'll notice:

• Being bounded means the function's output won't go to infinity.
• This property is vital for applying the Squeeze Theorem.

In our exercise, \( e^{\sin(\pi/x)} \) is bounded between \( e^{-1} \) and \( e^1 \). With this understanding, when multiplying by \( \sqrt{x} \), which trends towards zero, the combined effect allows us to use the Squeeze Theorem effectively. Since both bounding expressions approach zero, the product with \( \sqrt{x}e^{\sin(\pi/x)} \) becomes squeezed to a limit of 0 as well. This demonstrates how bounded functions can simplify the evaluation of complex limits.