Evaluate the following limits:
`lim_(xrarr0)((e^(2+x)-e^(2))/(x))`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^{2+x} - e^2}{x}, \]

we start by substituting \(x = 0\):

\[ \frac{e^{2+0} - e^2}{0} = \frac{e^2 - e^2}{0} = \frac{0}{0}. \]

This is an indeterminate form, so we can apply L'Hôpital's Rule, which states that if we have a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the derivative of the denominator.

1. **Differentiate the numerator**: The numerator is \(e^{2+x} - e^2\). The derivative of \(e^{2+x}\) with respect to \(x\) is \(e^{2+x}\) (since the derivative of \(e^u\) is \(e^u \cdot \frac{du}{dx}\) and \(u = 2+x\) gives \(\frac{du}{dx} = 1\)). The derivative of \(e^2\) (a constant) is \(0\). Thus, the derivative of the numerator is:

\[ \frac{d}{dx}(e^{2+x} - e^2) = e^{2+x} - 0 = e^{2+x}. \]

2. **Differentiate the denominator**: The denominator is simply \(x\), and its derivative is \(1\).

Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{e^{2+x}}{1}. \]

3. **Evaluate the limit**: Now we substitute \(x = 0\):

\[ e^{2+0} = e^2. \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{e^{2+x} - e^2}{x} = e^2. \]

**Final Answer**:

\[ \lim_{x \to 0} \frac{e^{2+x} - e^2}{x} = e^2. \]

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