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 differnetiable at x =0
C. continuous but not differentiable at x=0
D. Discontinuous at x=0

Answer

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

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

we need to check the limits as \( x \) approaches 0 from both sides and compare them to the value of the function at \( x = 0 \).

### Step 1: Find the left-hand limit as \( x \) approaches 0.
We calculate:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1 + e^{\frac{1}{x}}} \]

For \( x < 0 \), \( \frac{1}{x} \) approaches \( -\infty \) as \( x \) approaches 0 from the left. Therefore, \( e^{\frac{1}{x}} \) approaches \( e^{-\infty} = 0 \).

Thus, we have:

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

### Step 2: Find the right-hand limit as \( x \) approaches 0.
Next, we calculate:

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

For \( x > 0 \), \( \frac{1}{x} \) approaches \( +\infty \) as \( x \) approaches 0 from the right. Therefore, \( e^{\frac{1}{x}} \) approaches \( e^{+\infty} = \infty \).

Thus, we have:

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

### Step 3: Compare limits and the value of the function at \( x = 0 \).
Now, we compare the left-hand limit, right-hand limit, and the value of the function at \( x = 0 \):

- Left-hand limit: \( \lim_{x \to 0^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 0 \)
- Value of the function at \( x = 0 \): \( f(0) = 0 \)

Since the left-hand limit (1) is not equal to the right-hand limit (0), we conclude that:

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

### Conclusion:
The function \( f(x) \) is discontinuous at \( x = 0 \).

### Final Answer:
The function \( f(x) \) is discontinuous at \( x = 0 \).

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