Let `f(x)={{:(x^(n)sin(1//x^(2))","xne0),(0", "x=0):},(ninI).` Then

Options

A. `(pi)/(2)`
B. `(pi)/(2sqrt(2))`
C. `(pi)/(sqrt(2))`
D. `sqrt(2)pi`

Explanation

To find the limit of the function \( f(x) \) as \( x \) approaches 0, we have:

\[ f(x) = \begin{cases} x^n \sin\left(\frac{1}{x^2}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We need to evaluate:

\[ \lim_{x \to 0} f(x) \]

### Step 1: Analyze the function as \( x \) approaches 0

When \( x \) approaches 0, the term \( \sin\left(\frac{1}{x^2}\right) \) oscillates between -1 and 1. Therefore, we can say:

\[ -1 \leq \sin\left(\frac{1}{x^2}\right) \leq 1 \]

### Step 2: Multiply by \( x^n \)

Now, we can multiply the inequality by \( x^n \):

\[ -x^n \leq x^n \sin\left(\frac{1}{x^2}\right) \leq x^n \]

### Step 3: Consider the limit of the bounds

Next, we take the limit of the bounds as \( x \) approaches 0:

1. If \( n > 0 \):
\[ \lim_{x \to 0} -x^n = 0 \quad \text{and} \quad \lim_{x \to 0} x^n = 0 \]
By the Squeeze Theorem:
\[ \lim_{x \to 0} x^n \sin\left(\frac{1}{x^2}\right) = 0 \]

2. If \( n = 0 \):
\[ f(x) = \sin\left(\frac{1}{x^2}\right) \]
The limit does not exist because \( \sin\left(\frac{1}{x^2}\right) \) oscillates between -1 and 1.

3. If \( n < 0 \):
\[ -x^n \to \infty \quad \text{and} \quad x^n \to \infty \]
In this case, the limit does not exist as \( x \to 0 \).

### Conclusion

Thus, we conclude that:

\[ \lim_{x \to 0} f(x) = \begin{cases} 0 & \text{if } n > 0 \\ \text{does not exist} & \text{if } n = 0 \\ \text{does not exist} & \text{if } n < 0 \end{cases} \]