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. Use the Squeeze Theorem to show that \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0.\)

Short answer

Question: Prove the inequality \(-|x| \leq x \sin \frac{1}{x} \leq |x|\) for \(x \neq 0\), graph the inequalities, and then use the Squeeze Theorem to find the limit of \(x \sin \frac{1}{x}\) as \(x\) approaches 0. Answer: Proved inequality: \(-|x| \leq x \sin \frac{1}{x} \leq |x|\) for \(x \neq 0\). Graphing the inequalities, we observe that the graph of \(y = x \sin \frac{1}{x}\) is bounded by the graphs of \(y = -|x|\) and \(y = |x|\). By applying the Squeeze Theorem, we conclude that \(\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0\).

Step-by-step solution

Prove the inequality

First, we need to show that \(-|x| \leq x \sin \frac{1}{x} \leq |x|\) for \(x \neq 0\). We can break this down into two separate inequalities: 1) \(-|x| \leq x \sin \frac{1}{x}\) 2) \(x \sin \frac{1}{x} \leq |x|\) Let's start with inequality 1: By definition, the sine function has a range of \(-1 \leq \sin(x) \leq 1\). So, if we multiply both sides by \(-|x|\), we get: \(-|x| \leq -|x| \sin(x) \leq |x|\) Now, let \(h(x) = x \sin(x)\) for \(x \neq 0\). Then, \(-|x| \leq h(x) \leq |x|\) Thus, we have proved inequality 1. Moving on to inequality 2: As stated earlier, the sine function has a range of \(-1 \leq \sin(x) \leq 1\). So, if we multiply both sides by \(|x|\), we get: \(-|x| \leq |x| \sin(x) \leq |x|\) Now, let \(g(x) = x \sin(x)\): \(-|x| \leq g(x) \leq |x|\) Thus, we have proved inequality 2. So, combining both inequalities, we have: \(-|x| \leq x \sin \frac{1}{x} \leq |x|\) for \(x \neq 0\).

Graph the inequalities

In this step, we will illustrate the inequalities using a graph. Use a graphing tool like Desmos or Geogebra to plot the following functions: 1) \(y = -|x|\) 2) \(y = x \sin \frac{1}{x}\) 3) \(y = |x|\) You will see that the graph of \(y = x \sin \frac{1}{x}\) is bounded by the graphs of \(y = -|x|\) and \(y = |x|\).

Apply the Squeeze Theorem

The Squeeze Theorem states that if for certain functions \(f(x)\), \(g(x)\), and \(h(x)\), we have \(f(x) \leq g(x) \leq h(x)\), and the limits of \(f(x)\) and \(h(x)\) as \(x \rightarrow a\) are both L, then the limit of \(g(x)\) as \(x \rightarrow a\) also exists and is equal to L. From the inequalities we proved earlier, we have: \(f(x) = -|x|\) , \(g(x) = x \sin \frac{1}{x}\) , \(h(x) = |x|\) As \(x \rightarrow 0\), both \(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} -|x| = 0\) and \(\lim_{x \rightarrow 0} h(x) = \lim_{x \rightarrow 0} |x| = 0\). By the Squeeze Theorem: \(\lim_{x \rightarrow 0} g(x) = \lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0\) Therefore, \(\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0\).

Key concepts

Inequality Proof

To understand the inequality \(-|x| \leq x \sin \frac{1}{x} \leq |x|\) for \(x eq 0\), it's important to consider the basic properties of the sine function. The sine function oscillates between -1 and 1, given by the mathematical form \(-1 \leq \sin(x) \leq 1\). This property helps us bound the expressions involving sine.
When you multiply sine, \(\sin(x)\), by \(|x|\), the inequality transforms into \(-|x| \leq x \sin \frac{1}{x} \leq |x|\). Each side of the original inequality holds due to multiplying a non-negative piece, \(|x|\), with the sine function's output.
Understanding this proof involves recognizing how each part of \(x \sin \frac{1}{x}\) fits within the bounds marked by the functions \(-|x|\) and \(|x|\). Since the maximum and minimum possible values of \(\sin \frac{1}{x}\) are 1 and -1 respectively, scaling by \(|x|\) maintains these limits without violating the inequality conditions. This approach helps us confirm the given inequality for all \(x eq 0\).

Limits

In calculus, limits are pivotal in defining behaviors as variables approach specific values. The exercise uses the Squeeze Theorem involving limits to establish the behavior of the function \(x \sin \frac{1}{x}\) as \(x\) approaches 0. The theorem is a powerful tool when direct evaluation is complex or not possible.
To apply the Squeeze Theorem here, note that we already have \(-|x| \leq x \sin \frac{1}{x} \leq |x|\). If the limits of \(-|x|\) and \(|x|\) are identical as \(x\) approaches 0, then the limit of \(x \sin \frac{1}{x}\) must be the same due to its bounded nature.
- For \(-|x|\), as \(x\) approaches 0, \(|x|\) does too, causing \(-|x|\) to approach 0.
- Similarly, \(|x|\) itself approaches 0 as \(x\) approaches 0.
Therefore, both side limits are 0, allowing us to conclude via the Squeeze Theorem that \(\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0\). This demonstrates the handy nature of limits to predict function behavior when direct routes are blocked.

Graphical Illustration

Visualizing mathematical concepts is crucial for deep understanding, especially in inequalities like \(-|x| \leq x \sin \frac{1}{x} \leq |x|\). Graphical representations make this comprehension easier and more intuitive.
To graph these inequalities, consider plotting:
- The line \(y = -|x|\) which represents the lower bound
- The function \(y = x \sin \frac{1}{x}\) which oscillates intricately, bounded by the other two expressions
- The line \(y = |x|\) representing the upper bound
As you graph these functions, you'll observe that the curve of \(x \sin \frac{1}{x}\) energetically wiggles between the lines \(-|x|\) and \(|x|\). Despite being confined, it approaches the x-axis as \(x\) nears 0 from either direction. This illustration underpins not just the validity of the algebraic inequality but also exhibits the behavior visually, providing a coherent grasp of the Squeeze Theorem in action.