Question

a. Show that \(-|x| \leq x \sin \frac{1}{x} \leq|x|,\) for \(x \neq 0\). b. Illustrate the inequalities in part (a) with a graph. c. Show that \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0\).

Short answer

Question: Prove that \(-|x| \leq x \sin \frac{1}{x} \leq|x|\) for \(x \neq 0\), illustrate these inequalities in a graph, and prove that \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0\). Answer: We can prove the inequality \(-|x| \leq x \sin \frac{1}{x} \leq|x|\) for \(x \neq 0\) using the range of the sine function \(-1\leq \sin(\theta) \leq 1\). The inequality holds true because \(x\sin(\frac{1}{x})\) stays between \(-|x|\) and \(|x|\) for all \(x \neq 0\). To illustrate this, students can create a graph using graphing software like Desmos or GeoGebra, which should show the function oscillating between the boundary curves. Finally, we can use the Squeeze Theorem to prove that \(\lim_{x\rightarrow 0} x\sin(\frac{1}{x}) = 0\), as it is squeezed between the limits of \(-|x|\) and \(|x|\) both approaching 0 as \(x \rightarrow 0\).

Step-by-step solution

Part (a) - Proving Inequalities

To prove the inequality \(-|x| \leq x \sin \frac{1}{x} \leq|x|,\) for \(x \neq 0\), we will use the following facts: - \(\sin(\theta)\) is always in the range of \(-1\leq \sin(\theta) \leq 1\) for any value of \(\theta\). - \(|x|\) is the absolute value of \(x\), which is always non-negative. We know \(\sin \theta\) varies between \(-1\) and \(1\). Therefore, $$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$ Now, we multiply each inequality by the non-negative value \(|x|\) (note that \(x \neq 0\) by assumption): $$-|x|\leq |x|\sin\left(\frac{1}{x}\right) \leq |x|$$ We want to have the inequality in terms of \(x \sin \frac{1}{x}\): If \(x > 0\), then \(|x|=x\). Therefore, $$ -x \leq x\sin\left(\frac{1}{x}\right) \leq x$$ If \(x<0\), then \(|x|=-x\). Therefore, $$ -(-x) \leq -x \sin\left(\frac{1}{x}\right) \leq -x$$ or, $$ x \leq x \sin\left(\frac{1}{x}\right) \leq -x$$ Since the inequalities hold for \(x>0\) and \(x<0\), and we have \(x \neq 0\), the inequality holds for all \(x \neq 0\): $$-|x| \leq x \sin \frac{1}{x} \leq|x|$$

Part (b) - Graph Illustration

To illustrate the inequalities \(-|x| \leq x \sin \frac{1}{x} \leq|x|\) in a graph, first draw the functions \(y=x\sin(\frac{1}{x}), y=-|x|,\) and \(y=|x|\) on the same plot. The graph should show that \(x\sin(\frac{1}{x})\) is always between \(-|x|\) and \(|x|\). Note how the function oscillates between both boundary curves. Since this is a text-based explanation, we cannot provide a graph, but students can create a graph using graphing software like Desmos or GeoGebra.

Part (c) - Proving the Limit

To prove that \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0\), we can use the Squeeze Theorem. The Squeeze Theorem states that if a function \(g(x)\) is always between two other functions \(h(x)\) and \(f(x)\) such that \(h(x) \leq g(x) \leq f(x)\) and the limits of \(h(x)\) and \(f(x)\) as \(x\) approaches some point are the same, then the limit of \(g(x)\) at that point also equals the same value. From Part (a), we have: $$-|x| \leq x \sin \frac{1}{x} \leq|x|$$ Taking the limits of all three functions as \(x \rightarrow 0\), we get: $$\lim_{x \rightarrow 0} -|x| \leq \lim_{x \rightarrow 0} x \sin \frac{1}{x} \leq \lim_{x \rightarrow 0} |x|$$ Now, we know that the limits of \(-|x|\) and \(|x|\) as \(x\rightarrow 0\) are both 0. Therefore, $$0 \leq \lim_{x \rightarrow 0} x \sin \frac{1}{x} \leq 0$$ By applying the Squeeze Theorem, the limit of \(x\sin(\frac{1}{x})\) as \(x\rightarrow 0\) is also 0: $$\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$$

Key concepts

Squeeze Theorem

When you're exploring calculus and limits, the Squeeze Theorem can be a very useful tool. It's a method that helps in determining the limit of a function. Imagine having a function you're interested in, let's call it \( g(x) \), and it's trapped between two other functions, \( h(x) \) and \( f(x) \). The key is that if both \( h(x) \) and \( f(x) \) converge to the same limit as \( x \) approaches a certain point, then \( g(x) \) must also converge to that same limit.
For example, in our exercise, the function \( x \sin \left(\frac{1}{x}\right) \) is squeezed between \(-|x|\) and \(|x|\). Both of these functions approach 0 as \( x \) goes to 0. Thus, by the Squeeze Theorem, \( \lim_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0 \). This concept is very powerful when you're dealing with trigonometric functions or oscillations where direct evaluation is tricky.

Limits

Limits are central to calculus. They represent the value that a function approaches as the input approaches some value. For instance, the limit \( \lim_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0 \) explains that as \( x \) gets infinitesimally close to zero, the function \( x \sin \left(\frac{1}{x}\right) \) approaches the value 0.
Limits are essential in defining derivatives and integrals, the core components of calculus. They help us understand the behavior of functions at points where they might not be explicitly defined, or where they have discontinuities. It allows mathematicians to explore changes and trends in functions and can be applied in real-world scenarios like physics and engineering.

Inequalities

Inequalities are expressions of mathematical relationships involving greater than or less than signs. They help in comparing the magnitude or size of values. In the given exercise, the inequality \(-|x| \leq x \sin \left(\frac{1}{x}\right) \leq |x|\) is used.
This inequality derives from the property of the sine function, since \( \sin \theta \) is always between -1 and 1 for any real number \( \theta \). By multiplying this property with \( |x| \), we establish the boundaries for \( x \sin \left(\frac{1}{x}\right) \). This concept reinforces the structure and constraints that the given function has within its domain, and is crucial in utilizing the Squeeze Theorem to find limits.

Graphing Functions

Graphing functions offers visual understanding and insight into the behavior of mathematical functions. When looking at the graphs of \( y = x \sin \left(\frac{1}{x}\right) \), \( y = -|x| \), and \( y = |x| \), you'll see how \( x \sin \left(\frac{1}{x}\right) \) oscillates rapidly and always stays within the envelopes of \( -|x| \) and \( |x| \).
Visualizing these functions on a graph can illuminate why the Squeeze Theorem applies and how it confirms that \( x \sin \left(\frac{1}{x}\right) \) approaches 0 as \( x \) approaches 0. When you draw or view these graphs, it becomes clearer how the function behaves near the point of interest, often providing a more intuitive understanding than calculations alone. Using graphing tools like Desmos or GeoGebra can be very helpful for this purpose.