`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

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

at \( x = 0 \), we need to check the left-hand limit and the right-hand limit as \( x \) approaches 0, and see if they are equal to \( f(0) \).

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

The left-hand limit is given by:

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

As \( x \) approaches 0 from the left (negative values), \( \frac{1}{x} \) approaches \(-\infty\). Therefore, we can evaluate:

\[ e^{\frac{1}{x}} \to e^{-\infty} = 0 \]

Thus, we have:

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

### Step 2: Find the right-hand limit as \( x \) approaches 0

The right-hand limit is given by:

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

As \( x \) approaches 0 from the right (positive values), \( \frac{1}{x} \) approaches \(+\infty\). Therefore, we can evaluate:

\[ e^{\frac{1}{x}} \to e^{+\infty} = \infty \]

Thus, we have:

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

### Step 3: Compare the limits and the function value 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: \( \lim_{x \to 0^-} f(x) = 0 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 1 \)
- Function value: \( f(0) = 0 \)

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

\[ \text{The function } f(x) \text{ is not continuous at } x = 0. \]

### Final Conclusion

The function \( f(x) \) is not continuous at \( x = 0 \).

---