Let ` f(x) = sin"" 1/x, x ne 0` Then f(x) can be continuous at x =0

Options

A. If f(0) =1
B. If f(0) =0
C. If f(0) = -1
D. For no value of (0)

Answer

To determine whether the function \( f(x) = \sin\left(\frac{1}{x}\right) \) can be continuous at \( x = 0 \), we need to analyze the behavior of the function as \( x \) approaches 0.

### Step 1: Define Continuity
A function \( f(x) \) is continuous at a point \( c \) if the following three conditions are satisfied:
1. \( f(c) \) is defined.
2. \( \lim_{x \to c} f(x) \) exists.
3. \( \lim_{x \to c} f(x) = f(c) \).

In our case, we want to check if \( f(x) \) can be continuous at \( x = 0 \).

### Step 2: Check if \( f(0) \) is Defined
The function \( f(x) = \sin\left(\frac{1}{x}\right) \) is defined for all \( x \neq 0 \). We need to define \( f(0) \) in order to check continuity. We will consider different possible values for \( f(0) \) based on the options given.

### Step 3: Find the Limit as \( x \to 0 \)
We need to evaluate the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \]
As \( x \) approaches 0, \( \frac{1}{x} \) approaches \( \infty \) (from the right side) and \( -\infty \) (from the left side). Therefore, we need to consider the limits from both sides:

- As \( x \to 0^+ \):
\[ \lim_{x \to 0^+} \sin\left(\frac{1}{x}\right) \text{ oscillates between -1 and 1.} \]

- As \( x \to 0^- \):
\[ \lim_{x \to 0^-} \sin\left(\frac{1}{x}\right) \text{ also oscillates between -1 and 1.} \]

Since \( \sin\left(\frac{1}{x}\right) \) does not approach a single finite value as \( x \) approaches 0, the limit does not exist.

### Step 4: Conclusion
Since the limit \( \lim_{x \to 0} f(x) \) does not exist, we cannot satisfy the condition for continuity at \( x = 0 \). Therefore, \( f(x) \) cannot be continuous at \( x = 0 \) for any value of \( f(0) \).

### Final Answer
The correct conclusion is that for no value of \( f(0) \) can the function \( f(x) = \sin\left(\frac{1}{x}\right) \) be continuous at \( x = 0 \).