When trying to derive, from first principles, the fact that exponential functions $a^x$ (where $a>1$ is real) are differentiable, we easily see that $$ \lim_{h\to0} \frac{a^{x+h}-a^x}h = a^x \lim_{h\to0} \frac{a^h-1}h, $$ provided the latter limit exists. It's even pretty easy to see that $$ \lim_{h\to0} \frac{a^h-1}h = ( \log_b a ) \lim_{h\to0} \frac{b^h-1}h $$ for any other real $b>1$, provided the latter limit exists. (And then one can define $e$ to be the number such that $\lim_{h\to0} \frac{e^h-1}h = 1$ and continue.)
So my question, which doesn't seem to have an answer on this site (though I'd be happy to be proved wrong) nor in the textbooks I've consulted: how can one justify the existence of any limit of the form $\lim_{h\to0} \frac{b^h-1}h$ $(b>1)$, without using the as-yet-underived fact that $b^x$ is differentiable? (Edited to add: I also want to avoid infinite series.)
from Hot Weekly Questions - Mathematics Stack Exchange
Greg Martin
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