Limit of a ratio between exponential function and its variable? How do I solve this limit without using L'Hopital's rule, if at all possible?
$$ \lim_{h\to0}\frac{1 - 2^{-h}}{h}$$
 A: Observe
$$\lim_{h\to 0}\frac{1-2^{-h}}{h}=-\lim_{h\to 0}\frac{2^{-h}-2^{-0}}{h}=...$$
(Hint: Define the derivative of $2^{-x}$ at $0$)
EDIT: The definition of the derivative of $2^{-x}$ at $0$ is
$$\lim_{h\to 0}\frac{2^{-h}-2^{-0}}{h}$$
But that is $(2^{-x})^{\prime}(0)=(-\ln 2\cdot 2^{-x})(0)=-\ln 2\cdot 2^0=-\ln 2$
A: You can expand the Numerator as a power series. 
Using $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \dots$,
$$2^{-h} = e^{\log(2) \cdot (-h)} = 1 - \log(2)\cdot h + \frac{(\log(2)\cdot h)^2}{2} \dots$$
Therefore 
$$\lim_{h\to0}\frac{1 - 2^{-h}}{h} = \lim_{h\to0}\frac{\log(2)\cdot h - \frac{(\log(2)\cdot h)^2}{2} \dots}{h} = \lim_{h\to0} \left( \log(2) - \frac{(\log(2))^2h}{2} \dots \right) = \log(2)$$
A: Note that $$\lim\limits_{x\to{0}}{\left( 1+x \right)^{\tfrac{1}{x}}}=e$$ is equivalent (by continuity of $\log{}$ function) to 
$$\lim\limits_{x\to{0}}{\dfrac{\log_a(1+x)}{x}}=\log_a{e}=\dfrac{1}{\ln{a}}.\;\;(a>0,\; a\ne{1}) \tag{*}$$
Substitute $t=\log_a(1+x),$ then $1+x=a^t, \;\; x=a^t-1$ and rewrite  $({}^*)$ as 
$$\lim\limits_{t\to{0}}{\dfrac{t}{a^t-1}}=\dfrac{1}{\ln{a}}$$
or
$$\lim\limits_{t\to{0}}{\dfrac{a^t-1}{t}}=\ln{a}.$$
