The value of `lim_(xrarroo) (sinx)/(x)`, is

Options

A. 1
B. 0
C. `-1`
D. none of these

Answer

To find the value of the limit \( \lim_{x \to \infty} \frac{\sin x}{x} \), we can follow these steps:

### Step 1: Understand the behavior of \( \sin x \)
The function \( \sin x \) oscillates between -1 and 1 for all values of \( x \). Therefore, we can say:
\[ -1 \leq \sin x \leq 1 \]

### Step 2: Set up the limit
We are interested in the limit:
\[ \lim_{x \to \infty} \frac{\sin x}{x} \]

### Step 3: Apply the Squeeze Theorem
Since \( \sin x \) is bounded, we can use the bounds of \( \sin x \) to create inequalities:
\[ -1 \leq \sin x \leq 1 \]
Dividing the entire inequality by \( x \) (which is positive as \( x \to \infty \)):
\[ -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]

### Step 4: Evaluate the limits of the bounding functions
Now, we take the limit of the bounding functions as \( x \) approaches infinity:
\[ \lim_{x \to \infty} -\frac{1}{x} = 0 \]
\[ \lim_{x \to \infty} \frac{1}{x} = 0 \]

### Step 5: Apply the Squeeze Theorem
Since both bounding limits converge to 0, by the Squeeze Theorem, we conclude that:
\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{0} \]