How do I prove a "double limit"?

Question

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.

Answer 1

Let $f_b$ be the function defined by $f_b(x) = b^x$ . Then: $$\lim_{b \rightarrow \infty} \lim_{h \rightarrow 0} \frac{b^h - 1}{h} = \lim_{b \rightarrow \infty} f_b'(0) = \lim_{b \rightarrow \infty} \log(b) = \infty$$

Answer 2

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

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