How to evaluate the limit $\lim_{h \to 0} \frac{e^{2+h}-e^2}h$? $$ {\lim \limits_{h \to 0}} { {e^{2+h}-e^2 } \over {h} }  $$
Due to time constraints, evaluating limits with e in them wasn't covered and I have this on the AP exam review. How do I proceed?

At the beginning of the year, in your AP Calculus AB class, you found derivatives by using one of two definitions:
1)  The derivative of a function at a specific x-value:  $\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)$
2)  The derivative of a function "in general", i.e. the "derivative function":  $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)$
The second formula gives the derivative at a general $x$-value.  But, if you plug in a value for $x$ in the formula (such as $2$), then you can get a formula, similar to #1, that gives the derivative of a function at a specific $x$-value:
3)  $\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=f'(a)$
So, for your problem, notice that it is the form of the formula #3...your $f(x)=e^x$, and your $a$ value is $2$.  Then, formula #3 becomes:
$\lim_{h\rightarrow 0}\frac{f(2+h)-f(2)}{h}=f'(2)$
$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=f'(2)$
Note that, for $f(x)=e^x$, $f'(x)=e^x$, so:
$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=e^2$
