The value of f(1) so that f(x)=`(e^x-e)//(x-1)` is continuous at x =1 is

Options

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

Answer

To find the value of \( f(1) \) such that the function

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

is continuous at \( x = 1 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 1 is equal to \( f(1) \).

### Step-by-Step Solution:

1. **Identify the limit**: We need to compute

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

2. **Substitute \( x = 1 \)**: If we directly substitute \( x = 1 \):

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

This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

3. **Apply L'Hôpital's Rule**: According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \( e^x - e \) is \( e^x \).
- The derivative of the denominator \( x - 1 \) is \( 1 \).

Thus, we have:

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

4. **Evaluate the limit**: Now substituting \( x = 1 \):

\[ \lim_{x \to 1} e^x = e^1 = e. \]

5. **Set \( f(1) \)**: For \( f(x) \) to be continuous at \( x = 1 \), we need:

\[ f(1) = \lim_{x \to 1} f(x) = e. \]

### Final Answer:
Thus, the value of \( f(1) \) so that \( f(x) \) is continuous at \( x = 1 \) is

\[ \boxed{e}. \]