`Lt_(x rarr oo) x sin (3/x)` is equal to

Options

A. `(1)/(3)`
B. 3
C. 1
D. does not exist

Answer

To solve the limit \( \lim_{x \to \infty} x \sin\left(\frac{3}{x}\right) \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to \infty} x \sin\left(\frac{3}{x}\right) \]
As \( x \) approaches infinity, \( \frac{3}{x} \) approaches 0. Therefore, we can rewrite the limit as:
\[ \lim_{x \to \infty} x \sin\left(\frac{3}{x}\right) = \lim_{x \to \infty} \frac{\sin\left(\frac{3}{x}\right)}{\frac{1}{x}} \]

### Step 2: Identify the indeterminate form
As \( x \to \infty \), \( \sin\left(\frac{3}{x}\right) \to \sin(0) = 0 \) and \( \frac{1}{x} \to 0 \). Thus, we have the indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists. We differentiate the numerator and denominator:
- The derivative of \( \sin\left(\frac{3}{x}\right) \) using the chain rule is:
\[ \frac{d}{dx}\left(\sin\left(\frac{3}{x}\right)\right) = \cos\left(\frac{3}{x}\right) \cdot \left(-\frac{3}{x^2}\right) \]
- The derivative of \( \frac{1}{x} \) is:
\[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]

### Step 4: Substitute the derivatives back into the limit
Now we substitute these derivatives back into the limit:
\[ \lim_{x \to \infty} \frac{\cos\left(\frac{3}{x}\right) \cdot \left(-\frac{3}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \to \infty} 3 \cos\left(\frac{3}{x}\right) \]

### Step 5: Evaluate the limit
As \( x \to \infty \), \( \frac{3}{x} \to 0 \), so:
\[ \cos\left(\frac{3}{x}\right) \to \cos(0) = 1 \]
Thus, we have:
\[ \lim_{x \to \infty} 3 \cos\left(\frac{3}{x}\right) = 3 \cdot 1 = 3 \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \infty} x \sin\left(\frac{3}{x}\right) = 3 \]