How to calculate the limit using the fundamental limit (without L'Hopital's rule)

Question

How to calculate the limit:

$$\lim_{h\to 0}\frac{e^{\alpha h}-1}{e^{\beta h}-1}$$ with $\alpha, \beta\in \Bbb R$.

Using the following fundamental limit:

$$\lim_{h\to 0}\frac{e^{h}-1}{h}=1$$

I need to calculate it without using L'Hopital's rule.

Answer 1

On the suggestion of @Sean Roberson.

We know that

$$\lim_{f(h)\to 0}\frac{e^{f(h)}-1}{f(h)}=1$$

Now if $h\to 0$ also $\alpha h\to 0$. Same for $\beta$ i.e. $\beta h\to 0$. Multiplying and dividing by $\alpha\beta$ both of which are different from zero, we obtain: $$\lim_{h\to 0}\frac{e^{\alpha h}-1}{e^{\beta h}-1}\,\frac{\beta\alpha}{\alpha\beta}$$ and rearranging,

$$\lim_{\alpha h\to 0}\frac{e^{\alpha h}-1}{\alpha h}\cdot\lim_{\beta h\to 0}\frac{\beta h}{e^{\beta h}-1}\cdot\frac{\alpha}{\beta}=1\cdot 1\cdot \frac{\alpha}{\beta}=\frac{\alpha}{\beta}$$