Question

\(\lim_{x \to \infty} \frac{\sin x}{x} =\)

• A. 1
• B. -1
• C. \(\infty\)
• D. zero

Hint:

Whenever you encounter a limit of a bounded function divided by a function that approaches infinity, the overall limit will tend to zero. Here, \(\sin x\) is bounded between -1 and 1, and \(x\) approaches infinity.

Answer 1

The Correct Option is D

Step 1: Understand the behavior of the numerator and the denominator as \(x\) approaches infinity.
The numerator is \(\sin x\). The sine function oscillates between -1 and 1 for all real values of \(x\). Therefore, as \(x \to \infty\), \(\sin x\) does not approach a specific value but remains bounded within the interval \([-1, 1]\).
The denominator is \(x\). As \(x \to \infty\), the value of \(x\) becomes infinitely large.
Step 2: Apply the limit properties or the Squeeze Theorem.
We can use the Squeeze Theorem (also known as the Sandwich Theorem) to evaluate this limit. We know that for all real numbers \(x\), $$-1 \leq \sin x \leq 1$$ Divide all parts of the inequality by \(x\). Since we are considering the limit as \(x \to \infty\), \(x\) will be positive, so the inequality signs remain the same: $$-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$$ Now, let's consider the limits of the lower and upper bounds as \(x \to \infty\): $$\lim_{x \to \infty} -\frac{1}{x} = 0$$ $$\lim_{x \to \infty} \frac{1}{x} = 0$$ According to the Squeeze Theorem, if \(g(x) \leq f(x) \leq h(x)\) for all \(x\) in an interval containing \(a\) (except possibly at \(a\)), and if \(\lim_{x \to a} g(x) = L\) and \(\lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} f(x) = L\). In our case, \(g(x) = -\frac{1}{x}\), \(f(x) = \frac{\sin x}{x}\), \(h(x) = \frac{1}{x}\), and \(a = \infty\). We found that:
$$\lim_{x \to \infty} -\frac{1}{x} = 0$$
$$\lim_{x \to \infty} \frac{1}{x} = 0$$
Therefore, by the Squeeze Theorem:
$$\lim_{x \to \infty} \frac{\sin x}{x} = 0$$ Step 3: Conclude the value of the limit.
The limit of \(\frac{\sin x}{x}\) as \(x\) approaches infinity is zero.