The problem:
Find the limit as h→0 for (e^(2+h)-e^2) / h
I really have no idea how to do this.
As h->0, you get $\displaystyle \frac{0}{0}$
Use L'Hopital's rule
The derivative of the numerator is:
$\displaystyle \frac{d}{dh} (e^{2+h} -e^2) = \frac{d}{dh} (e^2 \times e^h - e^2) = e^2.e^h-0=e^{2+h}$
and the derivative of the denominator is $\displaystyle \frac{d}{dh} h = 1$
by L'hopital's rule , you then have:
$\displaystyle \lim_{h->0} \frac{e^{2+h}}{1}$
Well it's the slope of the tangent of the curve $\displaystyle y=e^x$ at the point $\displaystyle (0,1)$, that is: the limit of the slope of the secant with one fixed point in $\displaystyle (0,e^0)=(0,1)$ and the other point moving towards that fixed point ($\displaystyle h\to 0$).
Yes, this limit has the value 1.