If `f(x)` is continuous at `x=0`, where `f(x){:{((1)/(1+e^(1/x))", for " x!=0),(k", for " x=0):}`, then `k=`

Options

A. 1
B. 0
C. -1
D. does not exists

Answer

To find the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

The function is defined as:
\[ f(x) = \begin{cases} \frac{1}{1 + e^{1/x}} & \text{for } x \neq 0 \\ k & \text{for } x = 0 \end{cases} \]

### Step 1: Find the left-hand limit as \( x \) approaches 0
To find the left-hand limit, we consider \( x \) approaching 0 from the negative side (\( x \to 0^- \)):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1 + e^{1/x}} \]
As \( x \) approaches 0 from the left, \( \frac{1}{x} \) approaches \(-\infty\). Therefore, \( e^{1/x} \) approaches \( e^{-\infty} = 0 \).

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

### Step 2: Find the right-hand limit as \( x \) approaches 0
Next, we consider \( x \) approaching 0 from the positive side (\( x \to 0^+ \)):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1 + e^{1/x}} \]
As \( x \) approaches 0 from the right, \( \frac{1}{x} \) approaches \( +\infty \). Therefore, \( e^{1/x} \) approaches \( e^{+\infty} = +\infty \).

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

### Step 3: Set the limits equal to \( f(0) \)
For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit must equal the right-hand limit and both must equal \( f(0) \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = k \]
From our calculations:
- Left-hand limit: \( 1 \)
- Right-hand limit: \( 0 \)

Since the left-hand limit (1) is not equal to the right-hand limit (0), we conclude that \( k \) cannot be defined such that \( f(x) \) is continuous at \( x = 0 \).

### Conclusion
Thus, we find that \( k \) does not exist in a way that makes \( f(x) \) continuous at \( x = 0 \).