The value of k which makes the function defined by `f(x)={("sin"1/x,"if x"ne0),(k,"if x"=0):}`, continuous 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 (LHL), right-hand limit (RHL), and the function value at \( x = 0 \) are all equal.

### Step 1: Find the Right-Hand Limit (RHL) as \( x \to 0^+ \)

The right-hand limit is given by:

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

As \( x \) approaches 0 from the positive side, \( \frac{1}{x} \) approaches \( +\infty \). The sine function oscillates between -1 and 1, so:

\[ \lim_{x \to 0^+} \sin\left(\frac{1}{x}\right) \text{ does not exist, but oscillates between -1 and 1.} \]

### Step 2: Find the Left-Hand Limit (LHL) as \( x \to 0^- \)

Similarly, the left-hand limit is given by:

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

As \( x \) approaches 0 from the negative side, \( \frac{1}{x} \) approaches \( -\infty \). Again, the sine function oscillates between -1 and 1, so:

\[ \lim_{x \to 0^-} \sin\left(\frac{1}{x}\right) \text{ does not exist, but oscillates between -1 and 1.} \]

### Step 3: Set the Limits Equal to the Function Value at \( x = 0 \)

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

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

Since both limits oscillate between -1 and 1, we can conclude that \( k \) must also lie within this range to maintain continuity:

\[ k \in [-1, 1] \]

### Conclusion

The value of \( k \) which makes the function continuous at \( x = 0 \) can be any value in the interval:

\[ k \in [-1, 1] \]