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

Explanation

To check 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 verify the following conditions:

1. The left-hand limit as \( x \) approaches 0 exists.
2. The right-hand limit as \( x \) approaches 0 exists.
3. Both limits are equal to \( f(0) \).

### Step 1: Calculate the Left-Hand Limit

We start by calculating the left-hand limit:

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

As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \). Therefore, we can rewrite the limit as:

\[ \lim_{x \to 0^-} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}} = \lim_{h \to 0^+} \frac{e^{-\frac{1}{h}}}{1 + e^{-\frac{1}{h}}. \]

Since \( e^{-\frac{1}{h}} \to 0 \) as \( h \to 0^+ \), we have:

\[ \lim_{h \to 0^+} \frac{0}{1 + 0} = 0. \]

Thus,

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

### Step 2: Calculate the Right-Hand Limit

Next, we calculate the right-hand limit:

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

As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \). Therefore, we can rewrite the limit as:

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

Since \( e^{\frac{1}{h}} \to +\infty \) as \( h \to 0^+ \), we have:

\[ \lim_{h \to 0^+} \frac{+\infty}{1 + \infty} = 1. \]

Thus,

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

### Step 3: Evaluate \( f(0) \)

Now, we evaluate \( f(0) \):

\[ f(0) = 0. \]

### Conclusion

Now we compare the limits and the function value:

- 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 does not equal the right-hand limit, we conclude that:

\[ \text{The function } f(x) \text{ is discontinuous at } x = 0. \]