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Prove $$\lim_{b \to \infty} \lim_{h \to 0} \frac{b^h - 1}{h} = \infty$$

I have never worked with double limits before so I have no idea how to approach the problem. Please don't use "$e$" in your solutions, since the above limit is part of the derivation of "$e$", so for all purposes "$e$" hasn't been discovered yet.

I know absolutely no Calculus rules except for the very basics (power, chain, quotient etc.). I also know the squeeze theorem and intermediate value theorem.

Thanks.

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  • $\begingroup$ What exactly ARE we allowed to use? It can be solved with Taylor's Theorem and L'Hopital but I've got a sneaking feeling you don't want that. $\endgroup$ – Tyler Oct 15 '13 at 3:40
  • $\begingroup$ @Tyler I know absolutely no Calculus rules except for the very basics (power, chain, quotient etc.). I also know the squeeze theorem and intermediate value theorem. Thats pretty much it. $\endgroup$ – user100924 Oct 15 '13 at 3:41
  • $\begingroup$ What is the source of this problem? A textbook? $\endgroup$ – Tyler Oct 15 '13 at 3:48
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    $\begingroup$ @Tyler No, my professor was deriving "$e$" and he used a limit above without explaining why it worked. I was just curious. $\endgroup$ – user100924 Oct 15 '13 at 3:55
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Let $f_b$ be the function defined by $f_b(x) = b^x$ . Then: $$\lim_{b \rightarrow \infty} \lim_{h \rightarrow 0} \frac{b^h - 1}{h} = \lim_{b \rightarrow \infty} f_b'(0) = \lim_{b \rightarrow \infty} \log(b) = \infty$$

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  • $\begingroup$ The notation confused me. I don't understand what $x \rightarrow b^x$ means. $\endgroup$ – user100924 Oct 15 '13 at 3:57
  • $\begingroup$ I changed the notation. Now you also know what the the notation "The function defined by $x \mapsto b^x$" means :) $\endgroup$ – Arthur Oct 15 '13 at 4:04
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If we go in the order you wrote, we get

$$\lim_{h\to 0}\frac{b^h-1}h\stackrel{\text{l'Hospital}}=\lim_{h\to 0}b^h\log b=\log b$$

and now

$$\lim_{b\to\infty}\log b=\infty$$

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  • $\begingroup$ Sorry, I don't know what l'Hospital is. Can the limit be solved without using it? $\endgroup$ – user100924 Oct 15 '13 at 3:39
  • $\begingroup$ I could but I'm afraid that without $\;e\;$ , l'H, Taylor and all that it is going to be pretty hard to do it. I can't without something of this. $\endgroup$ – DonAntonio Oct 15 '13 at 3:42

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