`lim_(x to infty) (sin x)/x`

Options

A. 1
B. 0
C. `infty`
D. `-infty`

Answer

To solve the limit \( \lim_{x \to \infty} \frac{\sin x}{x} \), we can use the Squeeze Theorem (also known as the Sandwich Theorem). Here’s a step-by-step solution:

### Step 1: Identify the bounds for \( \sin x \)

We know that the sine function is bounded:
\[ -1 \leq \sin x \leq 1 \]

### Step 2: Divide the inequality by \( x \)

Since \( x \) is approaching infinity and is positive for large values, we can divide the entire inequality by \( x \) (keeping in mind that this does not change the direction of the inequalities):
\[ -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]

### Step 3: Take the limit of the bounds as \( x \to \infty \)

Now, we will take the limit of the lower and upper bounds as \( x \) approaches infinity:
\[ \lim_{x \to \infty} -\frac{1}{x} = 0 \]
\[ \lim_{x \to \infty} \frac{1}{x} = 0 \]

### Step 4: Apply the Squeeze Theorem

Since we have:
\[ -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]
and both the lower and upper limits approach 0, by the Squeeze Theorem, we conclude that:
\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]

### Final Answer

Thus, the limit is:
\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]

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