The value of `lim_(xrarr0) (x^2sin((1)/(x)))/(sinx)`, is

Options

A. 1
B. 0
C. `1//2`
D. none of these

Answer

To solve the limit \( \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} \]

### Step 2: Analyze the components
As \( x \to 0 \), \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, we can bound \( \sin\left(\frac{1}{x}\right) \) as follows:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]

### Step 3: Substitute the bounds into the limit
This gives us:
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

### Step 4: Divide by \(\sin x\)
Now, we can divide the entire inequality by \(\sin x\) (noting that \(\sin x \to 0\) as \(x \to 0\)):
\[ -\frac{x^2}{\sin x} \leq \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} \leq \frac{x^2}{\sin x} \]

### Step 5: Evaluate the limits of the bounding functions
Next, we need to find the limits of the bounding functions as \( x \to 0 \):
1. We know that \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \), hence:
\[ \lim_{x \to 0} \frac{x^2}{\sin x} = \lim_{x \to 0} x^2 \cdot \frac{1}{\sin x} = 0 \cdot 1 = 0 \]
2. Similarly, the negative bound:
\[ \lim_{x \to 0} -\frac{x^2}{\sin x} = 0 \]

### Step 6: Apply the Squeeze Theorem
Since both bounds approach 0, by the Squeeze Theorem, we conclude that:
\[ \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{\sin x} = 0 \]

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