Geometric intuition for why the slope of the tangent at the y-intercept of $y=b^x$ is $\ln(b)$? I understand how to prove that algebraically, but it's really amazing how the slope is exactly $\ln(b)$. My question is how can I develop a geometric feeling for it, or is it even possible to do so?
 A: If slope is $\frac {\Delta y}{\Delta x}=\frac {b^{x_1}-b^x}{x_1 - x}$ where $x_1$ is really really close to $x$.
If we let $x_h = x+h$ and $h$ is a really really small number then
slope is $\frac {b^{x+h} - b^{x}}{(x+h)-x} = \frac {b^xb^h -b^x}h= b^x \frac {b^h -1}h$
The value $\lim_{h\to 0^+} \frac {b^h-1}h$ is a constant value depending only on the value of $b$ and has nothing to do with $x$.

Now how surprised would you be if I told you $\lim_\limits{h\to 0^+} \frac {b^h-1}h = \ln b$?
.... okay, maybe that is surprising.
But consider let define the function as $\operatorname{BEATSME}(b) = \lim_\limits{h\to 0^+} \frac {b^h-1}h$.
We can not that for small values of $b$ that  $\operatorname{BEATSME}(b)$ is small and for large values of $b$ that $\operatorname{BEATSME}(b)$ is large.
So there must be some number where $\operatorname{BEATSME}(b) = 1$.
So lets define a  number $E$ to be  DEFINED to be the number where $\operatorname{BEATSME}(E) = 1$.
Would it surprise you if I told you that $E = e$ where $e$ is Euler's number?
Okay, most classes define $e$ as $\lim_\limits{n\to \infty} (1+\frac 1n)^n$.  Okay... if we define $e$ that way what is $\operatorname{BEATSME}(e)$?
$e=\lim_\limits{n\to\infty} (1+\frac 1n)^n = \lim_\limits{h\to \infty}(1 + h)^{\frac 1h}$.
And $\operatorname{BEATSME}(e) = \lim_{h\to 0} \frac {e^g-1}h = \lim_\limits{h\to \infty}\frac {(\lim_\limits{h\to \infty}(1 + h)^{\frac 1h})^h-1}h=$
$\lim_\limits{h\to 0^+} \frac {((1+h)^{\frac 1h})^h-1}h=$
$\lim _\limits{h\to 0^+}{(1+h)^1 - 1}h = \lim_\limits{h\to 0^+} \frac {(1+h)-1}h=\lim_\limits{h\to 0^+} \frac hh = 1$.
So that's it . $\operatorname{BEATSME}(e) = 1$.

Okay, but what is $\operatorname{BEATSME}(b)$ in general?
Okay  $\operatorname{BEATSME}(b) = \lim_\limits{h\to 0} \frac{b^h -1}h= \ln b$.
Why?
Let $e^K = b$.
Then $\lim_\limits{n\to \infty} (1+\frac 1n)^{nK} = b$.
Let $(1+\frac 1n)^{nK} =B$ for a very large $n$.
Then $(1+\frac 1n)^K = B^{\frac 1n}$
By binomial expansion $(1+\frac 1n)^K \approx 1 + K\frac 1n$ for  large $n$.
So $1+K\frac 1n = B^{\frac 1n}$
$K = n(B^{\frac 1n}-1)$.
Replace $n$ with $\frac 1h$ and we get $K = \frac {B^h-1}h$ and so the the 
$\lim_\limits{h\to 0^+} \frac {b^h -1}h = \ln h$.

Whew.  So the slope of the tangent line of $b^x$ is $b^x\lim_\limits{h\to 0} \frac {b^h-1}h = b^x\ln b$.
And at $x=0$ that is $\ln b$ exactly.
