The value of k which makes `f(x) = {{:(sin(1//x), x ne 0),(k, x =0):}`, continous at x =0 is:

Options

A. 8
B. 1
C. `-1`
D. none of these

Answer

To determine the value of \( k \) that makes the function

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

continuous at \( x = 0 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at \( x = 0 \) are all equal.

### Step 1: Find the Left-Hand Limit as \( x \) approaches 0
We compute the left-hand limit:

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

As \( x \) approaches 0 from the left, \( \frac{1}{x} \) approaches \(-\infty\). The sine function oscillates between -1 and 1 as its argument approaches \(-\infty\). Therefore, the left-hand limit does not exist in a traditional sense, but we can say:

\[ \lim_{x \to 0^-} f(x) \text{ oscillates between } -1 \text{ and } 1. \]

### Step 2: Find the Right-Hand Limit as \( x \) approaches 0
Now we compute the right-hand limit:

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

As \( x \) approaches 0 from the right, \( \frac{1}{x} \) approaches \( +\infty\). Similar to the left-hand limit, the sine function oscillates between -1 and 1. Thus:

\[ \lim_{x \to 0^+} f(x) \text{ oscillates between } -1 \text{ and } 1. \]

### Step 3: Set the Limits Equal to the Function Value at \( x = 0 \)
For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = k \]

However, since both limits oscillate between -1 and 1, we can conclude that for \( f(x) \) to be continuous at \( x = 0 \), \( k \) must also be within the range of -1 and 1.

### Conclusion
Since the left-hand limit and right-hand limit do not converge to a single value, we cannot find a specific value of \( k \) that makes the function continuous at \( x = 0 \). Therefore, the answer is:

\[ \text{None of these} \]