Problem 38 - Calculus Early Transcendentals

Question

Prove that $$\lim _{x \rightarrow 0^{+}} \sqrt{x} e^{\sin (\pi / x)}=0.$$

Step-by-Step Solution

Answer

The limit \( \lim _{x \to 0^{+}} \sqrt{x} e^{\sin (\pi / x)} = 0 \) by the Squeeze Theorem.

Step 1: Understand the Components of the Limit

The expression \( \sqrt{x} e^{\sin (\pi / x)} \) involves two parts: \( \sqrt{x} \), which goes to zero as \( x \to 0^+ \), and \( e^{\sin (\pi / x)} \), which oscillates between \( e^{-1} \) and \( e^1 \) because \( \sin(\pi/x) \) takes all values between -1 and 1 as \( x \to 0^+ \). The goal is to see how these behaviors impact the approach to zero.

Step 2: Analyze the Behavior of \( e^{\sin(\pi/x)} \)

The term \( \sin(\pi/x) \) oscillates between -1 and 1 as \( x \to 0^+ \). Consequently, \( e^{\sin(\pi/x)} \) oscillates between \( e^{-1} \) and \( e^{1} \). This means that \( e^{\sin(\pi/x)} \) is bounded because \[ \frac{1}{e} \leq e^{\sin(\pi/x)} \leq e. \]

Step 3: Use the Squeeze Theorem

Considering the expression \( \sqrt{x} e^{\sin(\pi/x)} \), we note \[ \sqrt{x} \cdot \frac{1}{e} \leq \sqrt{x} e^{\sin(\pi/x)} \leq \sqrt{x} \cdot e. \] As \( x \to 0^+ \), \( \sqrt{x} \to 0 \). Hence, both \( \sqrt{x}/e \) and \( \sqrt{x} \cdot e \) approach zero. By the Squeeze Theorem, the original expression must also approach zero.

Step 4: Conclusion of the Proof

Since both bounding expressions approach zero, \( \sqrt{x} e^{\sin(\pi/x)} \) must also approach zero from above, confirming that \[ \lim_{x \to 0^{+}} \sqrt{x} e^{\sin(\pi / x)} = 0. \] Therefore, the limit is proven to be zero.

Key Concepts

Squeeze Theorem

When faced with limits involving complicated expressions, like functions that oscillate or behave erratically, the Squeeze Theorem can be incredibly useful. The theorem applies when you have a function that fits between two other functions. If the two bounding functions approach the same limit, then the function in between must also approach that limit.

In our exercise, we apply the Squeeze Theorem to the expression \( \sqrt{x} e^{\sin(\pi/x)} \). Let's break it down:

• The function \( \sin(\pi/x) \) causes \( e^{\sin(\pi/x)} \) to oscillate between \( e^{-1} \) and \( e^1 \).
• As \( x \to 0^+ \), \( \sqrt{x} \) goes to zero.
• Therefore, \( \frac{\sqrt{x}}{e} \leq \sqrt{x} e^{\sin(\pi/x)} \leq \sqrt{x} \cdot e \).

Both \( \frac{\sqrt{x}}{e} \) and \( \sqrt{x} \cdot e \) approach zero as \( x \to 0^+ \). Hence, by the Squeeze Theorem, \( \sqrt{x} e^{\sin(\pi/x)} \) must also approach zero.

Oscillating Function

An oscillating function regularly changes its value up and down as it approaches a limit, never settling at a specific value. The peculiar nature of oscillating functions like \( \sin(\pi/x) \) is crucial in the given exercise. This particular sine function behaves unpredictably near zero because its input repeats over smaller and smaller intervals.

The oscillation of \( \sin(\pi/x) \) means:

• The values swing wildly between -1 and 1.
• Thus, \( e^{\sin(\pi/x)} \) takes on values between \( e^{-1} \) and \( e^1 \).
• This multiplying factor \( e^{\sin(\pi/x)} \) remains bounded within a safe range, contributing to the limit analysis.

The challenge with an oscillating function in limits is managing this unpredictability, which is why bounding it, as done in the Squeeze Theorem, is effective.

Exponential Function

Exponential functions are significant in mathematics for their consistent growth behavior. In the given limit problem, \( e^{\sin(\pi/x)} \) is central to our analysis. This function's _(e raised to the power of something)_ behavior determines how the oscillating inputs impact the limit approach.

Here’s how it functions:

• Even though \( \sin(\pi/x) \) oscillates wildly, the exponent remains within -1 and 1.
• Therefore, \( e^{\sin(\pi/x)} \) outputs values that don't increase indefinitely but stay between \( e^{-1} \) and \( e^1 \).

Despite \( \sin(\pi/x) \) acting chaotically, the exponential transformation keeps the value bounded, making \( e^{\sin(\pi/x)} \) more manageable in limits.

Continuity

Continuity is the smooth, unbroken transition of a function's value without sudden jumps or gaps. In understanding limits, a continuous function behaves predictably. Although \( e^{\sin(\pi/x)} \) isn't continuous due to its oscillation, understanding continuity helps frame why \( \sqrt{x} e^{\sin(\pi/x)} \) behaves as it does near zero.

Some key points:

• \( \sqrt{x} \) is continuous as \( x \to 0^+ \), smoothly transitioning towards zero.
• The oscillation in \( e^{\sin(\pi/x)} \) doesn't break this continuity but just shifts values within a fixed exponential range.

Thus, the bounded and continuous behavior of \( \sqrt{x} \) for positive \( x \) ensures that the entire expression \( \sqrt{x} e^{\sin(\pi/x)} \) will trend smoothly towards zero, despite the internal oscillation.

Boundary Behavior

Boundary behavior examines how functions behave as they approach a particular value, from one direction, typically nearing an asymptote or a limit boundary. In this problem, we study \( \sqrt{x} e^{\sin(\pi/x)} \) as \( x \to 0^+ \).

The behavior at this boundary is critical:

• \( \sqrt{x} \) rapidly decreases to zero, acting as a diminishing boundary-closer.
• Though \( \, e^{\sin(\pi/x)} \) is oscillating, it stays bounded and doesn't diverge.
• Thus, the boundary behavior simplifies; the rapidly decreasing \( \sqrt{x} \) ensures the expression heads smoothly to zero.

This examination highlights how boundaries in a limit open insights into convergence and the behavior of compound functions as one component approaches a defined value.