`f(x)={{:(e^(1//x)/(1+e^(1//x)),if x ne 0),(0,if x = 0):}` at `x = 0`

Explanation

To determine the continuity of the function \( f(x) \) at \( x = 0 \), we need to check if the following condition holds:

\[ \lim_{x \to 0} f(x) = f(0) \]

Given the function:

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

### Step 1: Calculate \( f(0) \)

From the definition of the function, we have:

\[ f(0) = 0 \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \):

\[ f(x) = \frac{e^{1/x}}{1 + e^{1/x}} \]

We need to evaluate the limit as \( x \) approaches 0.

### Step 3: Evaluate the limit

As \( x \to 0 \), \( \frac{1}{x} \to +\infty \) (since \( x \) is approaching 0 from the right) and \( e^{1/x} \to +\infty \). Thus, we can rewrite the limit:

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

To simplify, we can divide the numerator and the denominator by \( e^{1/x} \):

\[ = \lim_{x \to 0} \frac{1}{\frac{1}{e^{1/x}} + 1} \]

### Step 4: Analyze the limit

As \( x \to 0 \), \( e^{1/x} \to +\infty \), which implies \( \frac{1}{e^{1/x}} \to 0 \). Therefore, we have:

\[ = \frac{1}{0 + 1} = 1 \]

### Step 5: Compare the limit with \( f(0) \)

Now we compare the limit with \( f(0) \):

\[ \lim_{x \to 0} f(x) = 1 \quad \text{and} \quad f(0) = 0 \]

Since:

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

### Conclusion

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