Problem 65 - Thomas Calculus

Question

a. It can be shown that the inequalities $$1-\frac{x^{2}}{6}<\frac{x \sin x}{2-2 \cos x}<1$$ hold for all values of \(x\) close to zero. What, if anything, does this tell you about \begin{equation} \begin{array}{c}{\lim _{x \rightarrow 0} \frac{x \sin x}{2-2 \cos x} ?} \\\ \\\\{\text { Give reasons for your answer. }}\end{array} \end{equation} b. Graph \(y=1-\left(x^{2} / 6\right), y=(x \sin x) /(2-2 \cos x), \quad\) and \(y=1\) together for \(-2 \leq x \leq 2 .\) Comment on the behavior of the graphs as \(x \rightarrow 0\)

Step-by-Step Solution

Answer

\(\lim_{x \to 0} \frac{x \sin x}{2-2 \cos x} = 1\) due to the Squeeze Theorem. Graphically, \(\frac{x \sin x}{2-2 \cos x}\) approaches \(y = 1\) as \(x \rightarrow 0\).

Step 1: Understand L'Hôpital's Rule

When evaluating limits of functions that yield an indeterminate form like \(\frac{0}{0}\), L'Hôpital's Rule provides a way to find the limit by differentiating the numerator and the denominator separately. This rule applies directly when the original function approaches an indeterminate form.

Step 2: Consider the Given Inequalities

The inequalities \(1-\frac{x^{2}}{6}<\frac{x \sin x}{2-2 \cos x}<1\) suggest that the function \(f(x) = \frac{x \sin x}{2-2 \cos x}\) is squeezed between two approaching values near zero, thus, this indicates that it is possible to apply the Squeeze Theorem.

Step 3: Apply the Limit Using the Squeeze Theorem

According to the Squeeze Theorem, if you have \(g(x) < f(x) < h(x)\) and both \(g(x)\) and \(h(x)\) have the same limit \(L\) as \(x \to a\), then \(\lim_{{x \to a}} f(x) = L\). Since \(\lim_{{x \to 0}} ( 1 - \frac{x^{2}}{6} ) = 1\) and \(\lim_{{x \to 0}} 1 = 1\), it follows that \(\lim_{{x \to 0}} \frac{x \sin x}{2-2 \cos x} = 1\).

Step 4: Graph the Functions

Plot the graphs of \(y = 1-\frac{x^2}{6}\), \(y = \frac{x \sin x}{2-2 \cos x}\), and \(y = 1\) over the interval \(-2 \leq x \leq 2\). The graph will show that \(\frac{x\sin x}{2 - 2 \cos x}\) stays between the other two functions, and becomes nearly indistinguishable from \(y = 1\) near \(x = 0\).

Step 5: Analyze the Graphical Behavior

As \(x \rightarrow 0\), the graphs \(y = 1-\frac{x^2}{6}\) and \(y = 1\) converge to the same line \(y = 1\). The graph of \(y = \frac{x \sin x}{2-2 \cos x}\) approaches \(y = 1\) very closely, confirming the limit derived through the Squeeze Theorem.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a helpful tool for finding limits of functions that result in an indeterminate form such as \(\frac{0}{0}\). When faced with such a situation, you can apply the rule by differentiating the numerator and the denominator of your function separately. This means taking the derivative of the top part of your function and the derivative of the bottom part.

Importantly, L'Hôpital's Rule can only be used when the original limit is indeed an indeterminate form. After differentiating, you can try to evaluate the limit again. If you still end up with an indeterminate form, you might need to apply the rule multiple times.

For example, if you have a function \(f(x)\) that results in \(\frac{0}{0}\), first apply L'Hôpital's Rule. Then take the derivatives of the function parts and see if you can evaluate the limit. If the derivatives give a determinate form, your work is done, otherwise, repeat the process.

Graphical Analysis

Graphical analysis is a visual method to understand the behavior of functions. By plotting the functions on a graph, you can clearly see their values relative to each other over a given interval.

In the context of the original exercise, you're plotting the graphs of three functions:

• \(y = 1-\frac{x^2}{6}\)
• \(y = \frac{x \sin x}{2-2 \cos x}\)
• \(y = 1\)

Doing this helps you understand how the function \(\frac{x \sin x}{2-2 \cos x}\) behaves compared to the other two. At values of \(x\) close to zero, you can see all three lines approaching the same value, strengthening the analytical findings of the Squeeze Theorem.

For positive and negative values approaching zero, the graph shows that how closely \(\frac{x \sin x}{2-2 \cos x}\) is sandwiched between the other two functions. This convergence visually confirms the limit of the function as shown mathematically through analysis.

Limit Evaluation

Limit evaluation is a fundamental concept where you determine the value that a function approaches as the input draws closer to a certain point. In this context, it's about understanding what happens to \(\frac{x \sin x}{2-2 \cos x}\) as \(x\) approaches zero.

With the inequality given: \[1-\frac{x^{2}}{6}<\frac{x \sin x}{2-2 \cos x}<1\]we see that the function is squeezed, in other words, forced to stay between two simpler functions, \(1-\frac{x^{2}}{6}\) and a constant \(1\). As\( x \rightarrow 0\), both of these simpler functions also approach 1.

The Squeeze Theorem becomes your best friend here. If a function is caught between two other functions that both approach the same limit, the center function must also approach that limit. In this case, using the theorem allows us to conclude that \[ \lim_{x \rightarrow 0} \frac{x \sin x}{2-2 \cos x} = 1 \].
Understanding these foundational steps helps demystify why the function behaves that way and gives confidence to handle similar evaluations in the future.