If `f(x)={(e^((1)/(x))/(1+e^((1)/(x)))",", x ne0),(0",",x=0):}`, then

Options

A. `f(0^(+))=1`
B. `f(0^(+))=0`
C. `f(0^-))=1`
D. `f(0^(-))=0`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

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

We want to find the limits \( \lim_{x \to 0^+} f(x) \) and \( \lim_{x \to 0^-} f(x) \).

### Step 1: Calculate \( \lim_{x \to 0^+} f(x) \)

For \( x \to 0^+ \), \( \frac{1}{x} \) approaches \( +\infty \). Therefore, we can evaluate:

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

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

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

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

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

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

For \( x \to 0^- \), \( \frac{1}{x} \) approaches \( -\infty \). Thus, we evaluate:

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

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

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

### Step 3: Conclusion

We have found:

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

Since the left-hand limit and the right-hand limit are not equal, we conclude that \( f(x) \) is not continuous at \( x = 0 \).

### Final Answer

- \( \lim_{x \to 0^+} f(x) = 1 \)
- \( \lim_{x \to 0^-} f(x) = 0 \)
- \( f(x) \) is not continuous at \( x = 0 \).