The function `f(x)=x^(2)"sin"(1)/(x)`, `xne0,f(0)=0` at x=0

Options

A. is continuous but not differentiable
B. is discontinuous
C. is having discontinuous and differentiable
D. is continuous and differentiable

Explanation

To determine whether the function \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) for \( x \neq 0 \) and \( f(0) = 0 \) is continuous and differentiable at \( x = 0 \), we will follow these steps:

### Step 1: Check Continuity at \( x = 0 \)

**Definition of Continuity:**
A function \( f(x) \) is continuous at \( x = a \) if:
\[ \lim_{x \to a} f(x) = f(a) \]

Here, we need to check:
\[ \lim_{x \to 0} f(x) = f(0) \]

**Calculate \( f(0) \):**
\[ f(0) = 0 \]

**Calculate \( \lim_{x \to 0} f(x) \):**
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]

Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can use the Squeeze Theorem:
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

Taking the limit as \( x \to 0 \):
\[ \lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0 \]

Thus,
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

**Conclusion:**
\[ \lim_{x \to 0} f(x) = 0 = f(0) \]
Therefore, \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

**Definition of Differentiability:**
A function \( f(x) \) is differentiable at \( x = a \) if:
\[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \]

For \( a = 0 \):
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]

**Calculate \( f(h) \) when \( h \neq 0 \):**
\[ f(h) = h^2 \sin\left(\frac{1}{h}\right) \]

Thus,
\[ f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) \]

Using the fact that \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1:
\[ -h \leq h \sin\left(\frac{1}{h}\right) \leq h \]

Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} -h = 0 \quad \text{and} \quad \lim_{h \to 0} h = 0 \]

Thus,
\[ \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0 \]

**Conclusion:**
\[ f'(0) = 0 \]
So, \( f(x) \) is differentiable at \( x = 0 \).

### Final Conclusion:
The function \( f(x) \) is both continuous and differentiable at \( x = 0 \).