Let the function f(x) be defined as `f(x) = {:{((logx-1)/(x-e) , xnee),(k , x =e):}`
The value of k , for which the function is continuous at x =e , is equal to

Options

A. e
B. `1/e`
C. `e^(2)`
D. `-e`

Answer

To find the value of \( k \) for which the function \( f(x) \) is continuous at \( x = e \), we need to ensure that the left-hand limit and right-hand limit as \( x \) approaches \( e \) are equal to \( f(e) \).

Given the function:
\[ f(x) = \begin{cases} \frac{\log x - 1}{x - e} & \text{if } x \neq e \\ k & \text{if } x = e \end{cases} \]

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches \( e \)

We need to calculate:
\[ \lim_{x \to e} f(x) = \lim_{x \to e} \frac{\log x - 1}{x - e} \]

### Step 2: Substitute \( x = e \)

Substituting \( x = e \) directly into the function gives:
\[ f(e) = \frac{\log e - 1}{e - e} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator:
- The derivative of the numerator \( \log x - 1 \) is \( \frac{1}{x} \).
- The derivative of the denominator \( x - e \) is \( 1 \).

Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to e} \frac{\log x - 1}{x - e} = \lim_{x \to e} \frac{\frac{1}{x}}{1} = \lim_{x \to e} \frac{1}{x} \]

### Step 4: Evaluate the limit

Now substituting \( x = e \):
\[ \lim_{x \to e} \frac{1}{x} = \frac{1}{e} \]

### Step 5: Set the limit equal to \( k \)

For the function to be continuous at \( x = e \):
\[ k = \lim_{x \to e} f(x) = \frac{1}{e} \]

Thus, the value of \( k \) for which the function is continuous at \( x = e \) is:
\[ \boxed{\frac{1}{e}} \]