Evaluate the following limits :
`Lim _(x to 0 ) x sin (1/x)`

Answer

To evaluate the limit \( \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \]

### Step 2: Use the property of sine
We know that the sine function is bounded. Specifically, for any real number \( y \), we have:
\[ -1 \leq \sin(y) \leq 1 \]
Thus, for \( y = \frac{1}{x} \), we can write:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]

### Step 3: Multiply by \( x \)
Multiplying the entire inequality by \( x \) (noting that as \( x \to 0 \), \( x \) is approaching 0 from both sides), we get:
\[ -x \leq x \sin\left(\frac{1}{x}\right) \leq x \]

### Step 4: Analyze the limits of the bounding expressions
Now, we need to evaluate the limits of the bounding expressions as \( x \to 0 \):
\[ \lim_{x \to 0} -x = 0 \]
\[ \lim_{x \to 0} x = 0 \]

### Step 5: Apply the Squeeze Theorem
Since \( x \sin\left(\frac{1}{x}\right) \) is squeezed between \( -x \) and \( x \), and both of these limits approach 0, we can apply the Squeeze Theorem:
\[ \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0 \]

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