How to calculate this limit? Doesn't seem to be difficult, but still can't get it.
$\displaystyle\lim_{x\rightarrow 0}\frac{10^{-x}-1}{x}=\ln10$

It's worth noticing that we can define the natural logarithm as
$$\log x = \lim_{h \to 0} \frac{x^h-1}{h}$$
So in your case you have
$$ \lim_{h \to 0} \frac{10^{-h}-1}{h}= \lim_{h \to 0} \frac{\frac{1}{10}^{h}-1}{h}=-\log10$$
This result holds because we have that
$$ \lim_{h \to 0} \frac{x^h-1}{h} =\frac{0}{0}$$
So we can apply L'Hôpitals rule, differentiating with respect to $h$ to get
$$ \lim_{h \to 0} \frac{x^h-1}{h} =\lim_{h \to 0}  x^h \log x = \log x $$
Obivously this is done by knowing how to handle the derivative of an exponential function with arbitrary base $x$, so you could've also solved you problem by noticing the expression is a derivative, as other answers/comments suggest.
