Problem 69 - Calculus - AP Edition

Question

Find \(\lim _{x \rightarrow 0^{+}} \frac{x \sin (1 / x)}{\sin x}\) if it exists.

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Problem

We need to find the limit as \(x\) approaches 0 from the positive side of the expression \(\frac{x \sin (1/x)}{\sin x}\). This is potentially an indeterminate form, as both the numerator and denominator approach 0 when \(x\) approaches 0.

Step 2: Explore the Behavior of \(\sin(1/x)\)

As \(x\) approaches 0 from the positive side, \(1/x\) becomes very large, and thus \(\sin(1/x)\) oscillates between -1 and 1. However, this function is bounded, which will be useful in the limit computation.

Step 3: Apply the Squeeze Theorem

To use the Squeeze Theorem, note that since \(\sin(1/x)\) is bounded by -1 and 1, we have\[-x \leq x \sin(1/x) \leq x.\]Thus, dividing each part of the inequality by \(\sin x\), which is positive for \(x\) approaching 0 from the right,\[-\frac{x}{\sin x} \leq \frac{x\sin(1/x)}{\sin x} \leq \frac{x}{\sin x}.\]

Step 4: Evaluate \(\lim_{x\to0^+} \frac{x}{\sin x}\)

From trigonometric limits, we know \(\lim_{x\to0} \frac{x}{\sin x} = 1\). This tells us the following inequalities become:\[-1 \leq \lim_{x\to0^+} \frac{x\sin(1/x)}{\sin x} \leq 1.\]

Step 5: Conclude with the Squeeze Theorem

By the Squeeze Theorem, since the upper and lower limits bound the function to converge to the same value, we conclude:\(\lim_{x\to0^+} \frac{x\sin(1/x)}{\sin x} = 0.\)

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a comforting friend for taming limits with wild behavior. When you can't directly compute a limit, the Squeeze Theorem steps in. This theorem is especially handy when dealing with oscillating functions or functions that don't approach a clear value on their own. Here's the beauty of the method: if you can "trap" your function between two others that are both heading to the same limit, your function must follow.
For example, consider a function that oscillates but stays within a boundary like \(-1 \leq f(x) \leq 1\). If another function is always larger than \(-g(x)\) and smaller than \(g(x)\), and \(g(x)\) goes to zero as \(x\) approaches a point, then so must \(f(x)\).
In the exercise, we squeezed \((x \sin(1/x))/\sin x\) between \(-x/sin x\) and \(x/sin x\). Knowing these go to zero as \(x\) approaches zero makes the Squeeze Theorem give our answer confidently as \(0\). It’s a little like guiding a mischievous child through a crowded place—hold their hand on both sides, and you'll get them there safely!

Trigonometric Limits

Trigonometric limits are pivotal in calculus. They tame the unruly behavior of sine, cosine, and tangent functions around their key points. One of the most significant results you learn early on is \(\lim_{x\to0} \frac{\sin x}{x} = 1\). This adorable little formula is more powerful than it seems. It's like the secret weapon to simplifying several complex-looking limits.
In the exercise, we used this to evaluate \(\lim_{x\to0^+} \frac{x}{\sin x}\). By knowing how sine behaves at zero, we can figure out the fate of other limits that seem dizzying at first glance. Trigonometric limits reveal the underlying order hiding under apparent chaos. Often, when functions hover near undefined behavior, these basic limits help cut through the complexity.
By leveraging them, especially in concert with other rules like the Squeeze Theorem, you can navigate most limit problems with ease. Understand these thoroughly to have a fundamental toolbox brimming with insights!

Indeterminate Forms

Indeterminate forms are like the puzzles of calculus. They're expressions that brush against rules of division, multiplication, and powers but don’t follow them clearly. When we say something is an indeterminate form, we mean it seems to yield conflicting results or nonsensical forms, like \( \frac{0}{0}\), \(\frac{\infty}{\infty}\), or \(0 \cdot \infty\). These are peculiar because neither part of the expression can single-handedly determine the behavior as we approach the problem's critical point.
In the exercise, as \(x\) tends to zero, both the numerator and the denominator head towards zero. This zero-over-zero form is a classic indeterminate form that signals we need to dive deeper. Techniques like the Squeeze Theorem and limits help to articulate what this nebulous expression will become.
Recognizing indeterminate forms is crucial because they invite us to use calculus tools like L'Hôpital's Rule, algebraic simplification, or clever theorems (hello, Squeeze Theorem!) to resolve them properly. They transform what could be a frustrating dead-end into an exciting puzzle to solve. Mastering these scenarios opens up new insights across calculus problems.