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

Options

A. continuous at x =0
B. continuous and differnetiable at x =0
C. continuous but not differentiable at x=0
D. Discontinuous at x=0

Answer

To determine the continuity of the function \( f(x) \) defined as:

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

we need to check the limit of \( f(x) \) as \( x \) approaches 0 from both the left and the right.

### Step 1: Calculate the right-hand limit as \( x \) approaches 0

For \( x \to 0^+ \) (approaching from the right):

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

As \( x \) approaches 0 from the right, \( \frac{1}{x} \) approaches \( +\infty \). Therefore, \( e^{\frac{1}{x}} \) approaches \( +\infty \).

Thus,

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

So,

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

### Step 2: Calculate the left-hand limit as \( x \) approaches 0

For \( x \to 0^- \) (approaching from the left):

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

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

Thus,

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

So,

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

### Step 3: Compare the limits and the function value at \( x = 0 \)

Now we compare the limits we calculated:

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

Since the right-hand limit \( (0) \) and the left-hand limit \( (1) \) are not equal, we conclude that:

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

### Conclusion

Since the limit does not exist, the function \( f(x) \) is discontinuous at \( x = 0 \).

Thus, the answer is that \( f(x) \) is **discontinuous**.

---