How would I go about proving there exists a real number $ r $ such that: $$\lim_{x\rightarrow 0}\frac {r^x-1}{x}=1$$ and actually finding its value?
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You see that the limit is $f'(0)$, where $f(x) = r^x$? – Daniel Fischer May 06 '14 at 18:41
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2You know that the answer is $e$. How you prove it depends on the machinery that has been built up to this point in the course. – André Nicolas May 06 '14 at 18:43
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@AndréNicolas exactly. That's why I just asked this question. – user132181 May 06 '14 at 19:13
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1The thing to be very careful about is to make sure that the machinery we use does not essentially define $e$ as the number $r$ of your question. – André Nicolas May 06 '14 at 19:36
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Why not apply L'Hopital's Rule?
We know that, for all $r \neq 0$, we have $r^x-1 \to 0$ as $x \to 0$. Moreover, $x \to 0$ as $x \to 0$ (!)
The derivative of $r^x-1$, with respect to $x$, is $r^x \ln r$, while the derivative of $x$, w.r.t. $x$ is $1$. Hence
$$\lim_{x \to 0} \frac{r^x-1}{x} = \lim_{x \to 0}\frac{r^x \ln r}{1} = \ln r \left( \lim_{x \to 0} r^x\right)$$
If $r \neq 0$ then $r^x \to 1$ as $x \to 0$ and hence:
$$\lim_{x \to 0} \frac{r^x-1}{x} = \ln r$$
Finally, $\ln r = 1 \iff r = \mathrm{e}$, where $\mathrm{e}$ is the usual suspect: $\mathrm{e} \approx 2.718\ldots$
Fly by Night
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It's a wonderful answer, but it turns out that my question was ill-posed. The question I really wanted to get an answer to is this one. – user132181 May 06 '14 at 19:18