If `f(x)={(x^(k) sin((1)/(x))",",x ne 0),(0",", x =0):}` is continuous at x = 0, then

Options

A. `k in (-oo, 0)`
B. `k in (1, oo)`
C. `k in (-1, oo)`
D. None of these

Explanation

To determine the value of \( k \) for which the function

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

is continuous at \( x = 0 \), we need to analyze the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Understanding Continuity**:
A function \( f(x) \) is continuous at \( x = 0 \) if:
\[ \lim_{x \to 0} f(x) = f(0) \]
Here, \( f(0) = 0 \).

2. **Finding the Limit**:
We need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^k \sin\left(\frac{1}{x}\right) \]
Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can bound the limit:
\[ -|x^k| \leq x^k \sin\left(\frac{1}{x}\right) \leq |x^k| \]

3. **Applying the Squeeze Theorem**:
To use the Squeeze Theorem, we need to analyze the behavior of \( |x^k| \) as \( x \to 0 \):
- If \( k > 0 \), then \( |x^k| \to 0 \) as \( x \to 0 \).
- If \( k = 0 \), then \( |x^k| = 1 \) which does not approach 0.
- If \( k < 0 \), then \( |x^k| \to \infty \) as \( x \to 0 \).

Therefore, for the limit to equal 0, we must have \( k > 0 \).

4. **Conclusion**:
Thus, for \( f(x) \) to be continuous at \( x = 0 \), we require:
\[ k > 0 \]

### Final Answer:
The function \( f(x) \) is continuous at \( x = 0 \) if \( k > 0 \).

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