If `f(x)={((1)/(1+e^(1//x))"," ,x ne 0),(0",", x =0):}` then f(x) is :

Options

A. Continuous at x = 0
B. Continuous and differentiable at x = 0
C. Continuous but not differentiable at x = 0
D. Discontinuous at x = 0

Explanation

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{1}{1 + e^{\frac{1}{x}} & , x \neq 0 \\ 0 & , x = 0 \end{cases} \]

we need to analyze the behavior of \( f(x) \) as \( x \) approaches 0 from both the left and the right.

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

For \( x \) approaching 0 from the right (i.e., \( x \to 0^+ \)):
\[ f(x) = \frac{1}{1 + e^{\frac{1}{x}}} \]
As \( x \) approaches 0 from the positive side, \( \frac{1}{x} \) approaches \( +\infty \). Therefore, we have:
\[ e^{\frac{1}{x}} \to e^{+\infty} \to +\infty \]
This leads to:
\[ f(x) = \frac{1}{1 + \infty} = \frac{1}{\infty} = 0 \]
Thus, \( \lim_{x \to 0^+} f(x) = 0 \).

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

For \( x \) approaching 0 from the left (i.e., \( x \to 0^- \)):
\[ f(x) = \frac{1}{1 + e^{\frac{1}{x}}} \]
As \( x \) approaches 0 from the negative side, \( \frac{1}{x} \) approaches \( -\infty \). Therefore, we have:
\[ e^{\frac{1}{x}} \to e^{-\infty} \to 0 \]
This leads to:
\[ f(x) = \frac{1}{1 + 0} = 1 \]
Thus, \( \lim_{x \to 0^-} f(x) = 1 \).

### Step 3: Compare the Limits and the Function Value at \( x = 0 \)

Now we have:
- \( \lim_{x \to 0^+} f(x) = 0 \)
- \( \lim_{x \to 0^-} f(x) = 1 \)
- \( f(0) = 0 \)

Since the right-hand limit (0) does not equal the left-hand limit (1), we conclude that the limits do not match.

### Conclusion

Since the limits from the left and right do not equal each other, the function \( f(x) \) is discontinuous at \( x = 0 \).

Thus, the answer is that \( f(x) \) is discontinuous at \( x = 0 \).

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