If `f(x) = x sin (1//x), x ne 0`, then `lim_(x to 0) f(x)`=

Options

A. 1
B. 0
C. `-1`
D. not exist

Answer

To find the limit of the function \( f(x) = x \sin\left(\frac{1}{x}\right) \) as \( x \) approaches 0, we can follow these steps:

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

### Step 2: Change of Variable
Next, we can make a substitution to simplify our limit. Let \( h = \frac{1}{x} \). As \( x \to 0 \), \( h \to \infty \). Therefore, we can rewrite the limit as:
\[ \lim_{h \to \infty} \frac{\sin(h)}{h} \]

### Step 3: Analyze the Sine Function
We know that the sine function is bounded:
\[ -1 \leq \sin(h) \leq 1 \]
Dividing the entire inequality by \( h \) (which is positive as \( h \to \infty \)), we get:
\[ -\frac{1}{h} \leq \frac{\sin(h)}{h} \leq \frac{1}{h} \]

### Step 4: Apply the Squeeze Theorem
Now, we can take the limit of the bounds as \( h \to \infty \):
\[ \lim_{h \to \infty} -\frac{1}{h} = 0 \quad \text{and} \quad \lim_{h \to \infty} \frac{1}{h} = 0 \]
By the Squeeze Theorem, since \( \frac{\sin(h)}{h} \) is squeezed between two expressions that both approach 0, we conclude:
\[ \lim_{h \to \infty} \frac{\sin(h)}{h} = 0 \]

### Step 5: Conclusion
Thus, we find:
\[ \lim_{x \to 0} f(x) = 0 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} f(x) = 0 \]

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