Proving equation involving limits without derivatives

[V0ODO0CH1LD]

Homework Statement

This is not really a homework or a coursework question. But I got a warning that I should submit my post in this section of the website.. I'm just saying this because I don't know if the answer to my question is at all achievable. And if it is how I should go about trying to solve it. Anyway.. Here is my question:

Can I prove that:
$$\lim_{h \to 0} \frac{(\frac{a^h-1}{h})}{(\frac{b^h-1}{h})} = \lim_{h \to 0} \frac{a^h-1}{b^h-1} = log_ba$$
without using derivatives?

Of course the first equality is trivial But I thought it might be important, that's why it is there.

Homework Equations

Honestly don't know..

Maybe the squeeze theorem?

The Attempt at a Solution

I tried making both sides an exponent of b. Which would looks like:
$$b^{log_ba} = b^{\frac{a^h-1}{b^h-1}}$$
which means:
$$a = b^{\frac{a^h-1}{b^h-1}}$$
I don't even know if that is the right path though..

 

[Dick]

Homework Statement

This is not really a homework or a coursework question. But I got a warning that I should submit my post in this section of the website.. I'm just saying this because I don't know if the answer to my question is at all achievable. And if it is how I should go about trying to solve it. Anyway.. Here is my question:

Can I prove that:
$$\lim_{h \to 0} \frac{(\frac{a^h-1}{h})}{(\frac{b^h-1}{h})} = \lim_{h \to 0} \frac{a^h-1}{b^h-1} = log_ba$$
without using derivatives?

Of course the first equality is trivial But I thought it might be important, that's why it is there.

Homework Equations

Honestly don't know..

Maybe the squeeze theorem?

The Attempt at a Solution

I tried making both sides an exponent of b. Which would looks like:
$$b^{log_ba} = b^{\frac{a^h-1}{b^h-1}}$$
which means:
$$a = b^{\frac{a^h-1}{b^h-1}}$$
I don't even know if that is the right path though..

You could use the taylor series of a^h=exp(log(a)*h) and similarly for b. But, of course, the taylor series involves derivatives. I'm not really sure you can even define what a^h even means without using derivatives in some way. I'm not sure you should bother trying.

 

[lurflurf]

Derivatives are inherent to that, though they could be hidden.


$$\lim_{h \rightarrow 0} \frac{a^h-1}{h}=\log(a)$$

can be derived a number of ways and is analogous to

$$\lim_{h \rightarrow 0} (1+a h)^h=e^a$$