Question

Use the Squeeze Theorem, where appropriate, to evaluate the given limit. $$ \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) $$

Short answer

The limit is 0.

Step-by-step solution

Understand the limit expression

We need to evaluate \( \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) \). As \( x \rightarrow 0 \), \( \frac{1}{x} \rightarrow \infty \), meaning that \( \sin \left(\frac{1}{x}\right) \) oscillates between \(-1\) and \(1\). This indicates that the Squeeze Theorem can be applied.

Find boundary functions

Observe that \(-1 \leq \sin \left(\frac{1}{x}\right) \leq 1\) for all \( x eq 0 \). Multiplying the entire inequality by \( x \) gives \(-x \leq x \sin \left(\frac{1}{x}\right) \leq x\).

Evaluate limits of boundary functions

Now, let's compute the limit as \( x \to 0 \) of both boundary functions. We need \( \lim _{x \to 0} -x \) and \( \lim _{x \to 0} x \). Both limits equal 0, because as \( x \to 0 \), \(-x\) and \(x\) each approach 0.

Apply the Squeeze Theorem

The Squeeze Theorem states that if \(-x \leq x \sin \left(\frac{1}{x}\right) \leq x\) and \( \lim _{x \to 0} -x = \lim _{x \to 0} x = 0 \), then \( \lim _{x \to 0} x \sin \left(\frac{1}{x}\right) = 0 \).

Key concepts

Limits

Limits are a fundamental concept in calculus, crucial for understanding how functions behave as they approach a certain point. In simpler terms, the limit of a function as it approaches a particular value provides information about where the function is heading. For instance, if we consider our example, \( \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) \), the task is to find the value that the function approaches as \( x \) gets closer and closer to zero. Often, we use limits to identify the behavior of functions that might not be well-behaved or defined at that particular point.

When dealing with limits, it’s important to assess whether the function is continuous or not. If it’s not continuous, like our example with \( \sin \left(\frac{1}{x}\right) \), we must use techniques like the Squeeze Theorem to evaluate the limit. This technique helps us draw conclusions without requiring a straightforward evaluation by trapping the function between two other well-understood functions.

Trigonometric Functions

Trigonometric functions are those based on the ratios of a triangle's sides in a right angle triangle context, the most common ones being sine, cosine, and tangent. In our exercise, we come across \( \sin \left(\frac{1}{x}\right) \), a trigonometric function of particular interest. This function is intriguing because as \( x \) approaches zero, \( \frac{1}{x} \) goes to infinity. As a result, \( \sin \left(\frac{1}{x}\right) \) does not settle into a single value but rather continues to oscillate.

A property of the sine function is that it oscillates between -1 and 1, despite the transformation \( \sin \left(\frac{1}{x}\right) \). This property is crucial when applying the Squeeze Theorem, as it helps establish bounds for the more complex function. Understanding how trigonometric functions behave is integral to successfully applying tools like the Squeeze Theorem in calculus. In conclusion, the sine function is bounded, making it easier to apply constraints needed for the theorem.

Oscillating Functions

Oscillating functions like \( \sin \left(\frac{1}{x}\right) \) show repetitive fluctuations that do not settle into a single value. Their value continually changes, which can complicate efforts to find a limit. As \( x \to 0 \), \( \sin \left(\frac{1}{x}\right) \) oscillates between -1 and 1, reflecting the unending cycle characteristic of sine and other trigonometric functions.

These functions are a perfect candidate for the Squeeze Theorem because while they may not have clear limits at first glance, they can be "squeezed" between functions with known limits. For example, in the exercise, we establish that \( -x \leq x \sin \left(\frac{1}{x}\right) \leq x \). By determining that the limits of \( -x \) and \( x \) as \( x \to 0 \) are both zero, the Squeeze Theorem allows us to confirm that \( \lim _{x \to 0} x \sin \left(\frac{1}{x}\right) = 0 \). This highlights how oscillating functions, despite their wild behavior, can still be managed and understood through mathematical tools.