Evaluate the following limits :
`lim_(x to 2)(e^(x)-e^(2))/(x-2)`

Answer

To evaluate the limit

\[ \lim_{x \to 2} \frac{e^x - e^2}{x - 2}, \]

we first substitute \(x = 2\) into the expression:

1. **Substitution**:
\[ e^2 - e^2 = 0 \quad \text{and} \quad 2 - 2 = 0. \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

- **Numerator**: The derivative of \( e^x \) is \( e^x \), and the derivative of \( e^2 \) (a constant) is \( 0 \). Thus, the derivative of the numerator is:
\[ \frac{d}{dx}(e^x - e^2) = e^x. \]

- **Denominator**: The derivative of \( x - 2 \) is \( 1 \). Thus, the derivative of the denominator is:
\[ \frac{d}{dx}(x - 2) = 1. \]

3. **Re-evaluate the limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 2} \frac{e^x}{1}. \]

4. **Substituting \(x = 2\)**:
\[ = e^2. \]

Thus, the final answer is:
\[ \lim_{x \to 2} \frac{e^x - e^2}{x - 2} = e^2. \]