Problem 83 - Calculus: Early Transcendentals

Question

A sine limit It can be shown that \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1,\) for \(x\) near 0. a. Illustrate these inequalities with a graph. b. Use these inequalities to evaluate \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\).

Step-by-Step Solution

Answer

Question: Prove the inequalities \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1\) for \(x\) near 0 and find the limit \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\). Answer: By graphing the functions \(y_1(x) = 1-\frac{x^{2}}{6}\), \(y_2(x) = \frac{\sin x}{x}\), and \(y_3(x) = 1\), we can illustrate the inequalities \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1\) for \(x\) near 0. Using these inequalities, we can evaluate the limit \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) and find that it is equal to 1.

Step 1: Understanding the inequalities and graphing

To graph the inequalities \(1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1\), we need to plot three functions: \(y_1(x) = 1-\frac{x^{2}}{6}\), \(y_2(x) = \frac{\sin x}{x}\), and \(y_3(x) = 1\). We will use a graphing tool to draw these functions for x near 0.

Step 2: Evaluate the limit using inequalities

We are given the inequalities \(\frac{\sin x}{x} \geq 1-\frac{x^{2}}{6}\) and \(\frac{\sin x}{x} \leq 1\). As \(x \to 0\), it follows that both sides of these inequalities converge to 1, meaning that: $$\lim _{x \rightarrow 0} (1-\frac{x^{2}}{6}) \leq \lim _{x \rightarrow 0} \frac{\sin x}{x} \leq \lim _{x \rightarrow 0} 1$$ We can easily find these limits: $$\lim _{x \rightarrow 0} (1-\frac{x^{2}}{6}) = 1 - \lim _{x \rightarrow 0} \frac{x^{2}}{6} = 1$$ $$\lim _{x \rightarrow 0} 1 = 1$$ So, $$1 \leq \lim _{x \rightarrow 0} \frac{\sin x}{x} \leq 1$$ From these inequalities, we can conclude that: $$\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$$

Key Concepts

Trigonometric Limits

Trigonometric limits involve functions like sine, cosine, and tangent, which are crucial in calculus and analysis. The function \( \frac{\sin x}{x} \) is particularly interesting because it behaves nicely around certain values, particularly as \( x \to 0 \). This function is indeterminate at \( x = 0 \), but its limit can be evaluated using inequalities.

For example, using the Squeeze Theorem, we can find the limit when \( x \to 0 \) for \( \frac{\sin x}{x} \). By understanding that \( \sin x \approx x \) near 0, the limit can be precisely evaluated. The Squeeze Theorem applies here because we find that other functions closely bound \( \frac{\sin x}{x} \) above and below.

This theorem helps us conclude that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Knowing that trigonometric limits often involve such approximations, recognizing patterns in simple identities can make solving similar limits easier.

Graphing Functions

Graphs provide a visual representation of how functions behave. In this problem, we can illustrate the inequalities \( 1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1 \) with a graph. By plotting the boundary functions \( y_1(x) = 1-\frac{x^{2}}{6} \) and \( y_3(x) = 1 \), along with the target function \( y_2(x) = \frac{\sin x}{x} \), you can visually validate the inequalities.

Here's how to proceed:

• Plot \( y_1(x) \) and notice it curves slightly downwards due to \( -\frac{x^{2}}{6} \).
• Plot \( y_3(x) \), which is simply a horizontal line at \( y = 1 \).
• Then plot \( y_2(x) = \frac{\sin x}{x} \). Observe that it fits between \( y_1(x) \) and \( y_3(x) \) near 0.

Graphing these functions visualizes that \( \frac{\sin x}{x} \) gets "squeezed" towards 1, reinforcing the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This insight emphasizes how understanding graph behaviors aids in solving limits and confirming analytical conclusions.

Evaluating Limits

Evaluating limits is crucial for understanding the behavior of functions as they approach specific points. The Squeeze Theorem is a method to find these limits by "squeezing" a target function between two other functions that converge to the same value. In our problem, the inequalities \( 1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1 \) let us apply this theorem.

To apply the Squeeze Theorem:

• Identify bounding functions that approach the same limit as the target function.
• Confirm the conditions where \( y_1(x) \leq y_2(x) \leq y_3(x) \) hold true as \( x \to 0 \).

In this case:
\[ \lim_{x \to 0} (1-\frac{x^{2}}{6}) = 1 \] and \[ \lim_{x \to 0} 1 = 1 \]
Both converge to the same limit, hence by the Squeeze Theorem, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Understanding this method strengthens your ability to evaluate complex limits by considering the functions' behaviors around certain points. This skill is essential for advanced calculus.