The value of k which makes `f(x)={:{(sin(1/x)", for " x!=0),(k", for " x=0):}` continuous at `x=0` is

Options

A. 0
B. 1
C. -1
D. no value of k

Answer

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

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

continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

### Step-by-Step Solution:

1. **Understand the definition of continuity**: A function \( f(x) \) is continuous at \( x = a \) if:
\[ \lim_{x \to a} f(x) = f(a) \]
In our case, we want:
\[ \lim_{x \to 0} f(x) = f(0) = k \]

2. **Calculate the limit as \( x \) approaches 0**:
We need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \]

3. **Analyze the behavior of \( \sin\left(\frac{1}{x}\right) \)**:
The function \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \) approaches 0. This means that the limit does not settle at a single value.

4. **Use subsequences to show the limit does not exist**:
- Consider the subsequence where \( x = \frac{1}{2\pi n} \) for \( n \in \mathbb{Z}^+ \):
\[ \lim_{n \to \infty} \sin\left(2\pi n\right) = 0 \]
- Now consider another subsequence where \( x = \frac{1}{2\pi n + \frac{1}{2}} \):
\[ \lim_{n \to \infty} \sin\left(2\pi n + \frac{1}{2}\right) = 1 \]

5. **Conclusion on the limit**:
Since we have found two different subsequential limits (0 and 1), we conclude that:
\[ \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \text{ does not exist.} \]

6. **Final conclusion about \( k \)**:
Since the limit does not exist, there is no value of \( k \) that can make \( f(x) \) continuous at \( x = 0 \). Therefore, the answer is:
\[ \text{No value of } k. \]