Let `f(x)={{:(,x^(n)sin\ (1)/(x),x ne 0),(,0,x=0):}` Then f(x) is continuous but not differentiable at `x=0`. If

Options

A. `0 lt p le 1`
B. `1 le p lt oo`
C. `-oo lt p lt 0`
D. p=0

Explanation

To solve the problem, we need to analyze the function defined as:

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

We need to find the conditions on \( n \) such that \( f(x) \) is continuous at \( x = 0 \) but not differentiable at \( x = 0 \).

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

For \( f(x) \) to be continuous at \( x = 0 \), we need:

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

Calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^n \sin\left(\frac{1}{x}\right) \]

Since \( \sin\left(\frac{1}{x}\right) \) is bounded between -1 and 1, we have:

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

As \( x \to 0 \), \( x^n \to 0 \) if \( n > 0 \). Therefore:

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

Thus, \( \lim_{x \to 0} f(x) = 0 = f(0) \) if \( n > 0 \). Hence, \( f(x) \) is continuous at \( x = 0 \) for \( n > 0 \).

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

To check differentiability at \( x = 0 \), we need to find the derivative:

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

For \( h \neq 0 \):

\[ f(h) = h^n \sin\left(\frac{1}{h}\right) \]

Thus,

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

Since \( \sin\left(\frac{1}{h}\right) \) is bounded, we can analyze the limit:

\[ -\lvert h^{n-1} \rvert \leq h^{n-1} \sin\left(\frac{1}{h}\right) \leq \lvert h^{n-1} \rvert \]

- If \( n > 1 \), then \( h^{n-1} \to 0 \) as \( h \to 0 \), and \( f'(0) = 0 \).
- If \( n = 1 \), then \( f'(0) \) does not exist because \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1.
- If \( n < 1 \), then \( h^{n-1} \to \infty \) as \( h \to 0 \), and \( f'(0) \) does not exist.

### Conclusion

For \( f(x) \) to be continuous at \( x = 0 \) but not differentiable at \( x = 0 \):

- \( n \) must be greater than 0 for continuity.
- \( n \) must be less than or equal to 1 for non-differentiability.

Thus, the condition on \( n \) is:

\[ 0 < n < 1 \]