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?
 A: Hint. One may recall that for a differentiable function around $x_o$ we have 
$$
\frac{f(x_0+h)-f(x_0) }{h} \to f'(x_0),\quad h\to 0.
$$ Apply it to $f(x)=e^x$.
A: 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:
$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=e^2$
A: Hint: If $f(x)=e^x$, then the given limit is $f'(2)$.
A: $$\lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=e^2\cdot\lim_{h\to 0}\frac{e^h-1}{h}=e^2\cdot 1= e^2$$
This follows from the fact that,
$$e^x=1+x+\mathcal{O}(x^2)\implies e^x-1=x+\mathcal{O}(x^2)\implies \frac{e^x-1}{x}=1+\mathcal{O}(x)$$
A: Let's see what it boils down to. You can rewrite it as:
$$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\rightarrow 0}\frac{e^{2}e^{h}-e^2}{h}=e^2\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}.$$
So you only need to figure out this last limit. If you are allowed to make use of knowledge about the derivative of $e^x$, then you are done, because the last limit is this derivative evaluated at $0$.
If you are not allowed to do that, I suggest using a Taylor series expansion of $e^x$. Have you covered that yet?
A: $${\lim \limits_{h \to 0}} { {e^{2+h}-e^2 } \over {h} } = \frac{d}{dx}\left(e^x\right)\Big|_{x=2}.$$
the left hand side is the limit definition of the derivative of $e^x$ at $x = 2$, then right hand side. use $(e^x)' = e^x.$
