What is the value of `lim_(xto0)x^(2)"sin"((1)/(x))`?

Options

A. 0
B. 1
C. `1//2`
D. Limit does not exist

Answer

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

### Step 1: Analyze the function
We need to evaluate the limit as \( x \) approaches 0. The function \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 for all \( x \neq 0 \). Therefore, we can use this property to help us find the limit.

### Step 2: Apply the Squeeze Theorem
Since \( \sin\left(\frac{1}{x}\right) \) is bounded, we know:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]
Multiplying the entire inequality by \( x^2 \) (which is non-negative as \( x \) approaches 0), we get:
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

### Step 3: Take the limit of the bounding functions
Now, we take the limit of the bounding functions as \( x \) approaches 0:
\[ \lim_{x \to 0} -x^2 = 0 \]
\[ \lim_{x \to 0} x^2 = 0 \]

### Step 4: Apply the Squeeze Theorem
By the Squeeze Theorem, since \( x^2 \sin\left(\frac{1}{x}\right) \) is squeezed between \( -x^2 \) and \( x^2 \), both of which approach 0, we conclude:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

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