Question

question 10
(limlimits_{x \to infty} \frac{sin x}{x})
(circ pi/2)
(circ 1)
(circ \text{dne})
(circ 0)

Explanation:

Step1: Recall the range of sine function

The sine function \( \sin x \) has a range of \( [-1, 1] \), so \( -1 \leq \sin x \leq 1 \) for all real numbers \( x \).

Step2: Analyze the limit as \( x \to \infty \)

We have the function \( \frac{\sin x}{x} \). Dividing each part of the inequality for \( \sin x \) by \( x \) (assuming \( x>0 \); for \( x<0 \), the inequality signs will flip, but the absolute value analysis still holds), we get \( \frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \) when \( x > 0 \). Now, take the limit as \( x \to \infty \) of all three parts. We know that \( \lim_{x \to \infty} \frac{-1}{x} = 0 \) and \( \lim_{x \to \infty} \frac{1}{x} = 0 \). By the Squeeze Theorem, since \( \frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \) and both the left - hand and right - hand limits are 0, then \( \lim_{x \to \infty} \frac{\sin x}{x}=0 \).

Answer:

\( 0 \) (corresponding to the option "0")