Evaluate `lim_(x to oo) (sinx^(0))/(x).`

Explanation

To evaluate the limit \( \lim_{x \to \infty} \frac{\sin(x^\circ)}{x} \), we can follow these steps:

### Step 1: Understand the function
The function we are dealing with is \( \frac{\sin(x^\circ)}{x} \). Here, \( x^\circ \) indicates that \( x \) is in degrees.

### Step 2: Analyze the sine function
The sine function, \( \sin(x^\circ) \), oscillates between -1 and 1 for all values of \( x \). Therefore, we can say:
\[ -1 \leq \sin(x^\circ) \leq 1 \]

### Step 3: Set up the limit
Now, we can rewrite our limit using the bounds of the sine function:
\[ \lim_{x \to \infty} \frac{-1}{x} \leq \lim_{x \to \infty} \frac{\sin(x^\circ)}{x} \leq \lim_{x \to \infty} \frac{1}{x} \]

### Step 4: Evaluate the limits of the bounding functions
Next, we will evaluate the limits of the bounding functions as \( x \) approaches infinity:
\[ \lim_{x \to \infty} \frac{-1}{x} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{1}{x} = 0 \]

### Step 5: Apply the Squeeze Theorem
Since both bounding limits approach 0, we can apply the Squeeze Theorem:
\[ \lim_{x \to \infty} \frac{\sin(x^\circ)}{x} = 0 \]

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