`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 analyze the function defined as:

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

### Step 1: Find \( f(0) \)
By definition, when \( x = 0 \):
\[ f(0) = 0 \]

### Step 2: Find \( \lim_{x \to 0} f(x) \)
We need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 from both sides (left and right). Since \( f(x) \) is defined differently for \( x = 0 \), we only consider the case when \( x \neq 0 \):

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

### Step 3: Analyze the limit as \( x \to 0 \)
As \( x \) approaches 0 from the right (i.e., \( x \to 0^+ \)), \( \frac{1}{x} \) approaches \( +\infty \). Therefore, we can evaluate the limit:

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

As \( x \) approaches 0 from the left (i.e., \( x \to 0^- \)), \( \frac{1}{x} \) approaches \( -\infty \):

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

### Step 4: Conclude the limit
Since the left-hand limit and the right-hand limit are not equal:

\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]

### Step 5: Check continuity
For \( f(x) \) to be continuous at \( x = 0 \), the following must hold:

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

We found that:

\[ \lim_{x \to 0} f(x) \text{ does not exist (as it approaches 1 from the right and 0 from the left)} \]
and
\[ f(0) = 0 \]

Since \( \lim_{x \to 0} f(x) \neq f(0) \), the function \( f(x) \) is discontinuous at \( x = 0 \).

### Final Conclusion
Thus, the function \( f(x) \) is discontinuous at \( x = 0 \).

---