Historical Approach to $\lim_{x \to 0} \frac{e^{\alpha x} - e^{\beta x}}{x}$, without L'Hospital's Rule I encountered this problem, amongst others, in the slightly older Calculus textbook Piskunov's Differential and Integral Calculus when I was working with a student:
Calculate the limit
$$
\lim_{x \to 0} \frac{ e^{\alpha x} - e^{\beta x} }{x}.
$$
Now, in the textbook, this came as a problem directly after the section/chapter on limits and continuity, additionally where $e$ and $\ln$ were defined and were equivalence of order of infinitesimals was discussed (as I said, a slightly older text). Importantly, this was before the chapter on derivatives, so, however obvious in retrospect, techniques like L'Hospital's method and Taylor series where clearly not meant to be used. Furthermore, in this section, the textbook had defined $e$ as the number satisfying $\lim_{x \to \infty} (1 + \frac{1}{x})^x$ (Showing how this was equivalent to and generalized from $\lim_{n \to \infty} (1 + \frac{1}{n})^n$). So, the fact that $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ was also not available (sometimes, calculus texts opt to define $e$ as the unique real number satisfying the last limit). Given these restrictions, I imagine there must be some older techniques that were used to solve such limits as this, and other similar ones from the section, such as $\lim_{n \to \infty} n \cdot [ a^{\frac{1}{n}} - 1 ]$. I would be interested in what these techniques are - I imagine they are probably related to showing certain infinitesimals are of equivalent order, and have actually seen some slightly similar applications of that theory, but am nowhere near familiar enough with it to see how to apply it here, at least, not without looking into it further.
To be clear, I don't need assistance in solving the limit using "modern" methods, by which I suppose I mean techniques involving applications of derivatives, use of taylor series, $O(x)$, and other such approaches as I said above (although, if there is an approach I haven't mentioned that you are aware of, please feel free to share, as I would be interested in more ways to think about such limits!), but rather I am interested in the classical or historical techniques used to compute nontrivial limits such as the above, that I imagine would be what Piskunov had in mind for the student using his text. I suppose a similar kind of limit might be 
$$
\lim_{x \to 0} \frac{ \sin \tan x - \tan \sin x }{ \arcsin \arctan x - \arctan \arcsin x },
$$
a limit from Arnold's Trivium (https://www.math.upenn.edu/Arnold/Arnold-Trivium-1991.pdf) that has been discussed a few times in various places -although I am too lazy to find the links - for which a more sophisticated approach than the usual L'Hospital is in order.
(actually, I chased down a link to some discussion of the above: https://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold)
 A: This is not the type of solution I was looking for primarily, but following a helpful pointer from Winther, the limit follows rather trivially using the binomial expansion as follows:
$$
\frac{ e^{\alpha x} - e^{\beta x} }{x}
= \lim_{n \to \infty} \frac{1}{x} \left\{ \left( 1 + \frac{\alpha x}{n} \right)^n - \left( 1 + \frac{\beta x}{n} \right)^n \right\}
= \lim_{n \to \infty} \frac{1}{x} \sum_{k = 1} ^n \binom{n}{k} \left[ \left( \frac{\alpha x}{n} \right)^k - \left( \frac{\beta x}{n} \right)^k \right],
$$
as for $k=0$, the term in the sum is $0$, and so
$$
\lim_{x \to 0} \frac{ e^{\alpha x} - e^{\beta x} }{x}
=
\lim_{x \to 0} \lim_{n \to \infty} \frac{1}{x} \sum_{k = 1} ^n \binom{n}{k} \left[ \left( \frac{\alpha x}{n} \right)^k - \left( \frac{\beta x}{n} \right)^k \right]
$$
$$
=
\lim_{x \to 0} \lim_{n \to \infty} \sum_{k = 0} ^n \binom{n}{k+1} \frac{1}{n^{k+1}} \left[ \alpha \left( \alpha x \right)^k - \beta \left( \beta x \right)^k \right]
$$
and hence 
$$
\lim_{x \to 0} \frac{ e^{\alpha x} - e^{\beta x} }{x} 
= 
\alpha - \beta + \lim_{x \to 0} x \cdot \left\{ \lim_{n \to \infty}  \sum_{k = 0} ^n \binom{n}{k+2} \frac{1}{n^{k+2}} \left[ \alpha^{k+2} x^{k} - \beta^{k+2} x^{k} \right] \right\}
\to \alpha - \beta + 0.
$$
A: It can also be easily calculated by using series expansion as follows , $$\lim_{x\to 0}\frac{e^{\alpha x}-e^{\beta x}}{x}$$ $$=\lim_{x\to 0}\frac{\left(1+\frac{(\alpha x)}{1!}+\frac{(\alpha x)^2}{2!}+\ldots\right)-\left(1+\frac{(\beta x)}{1!}+\frac{(\beta x)^2}{2!}+\ldots\right)}{x}$$
$$=\lim_{x\to 0}\frac{\left(\frac{(\alpha x)}{1!}+\frac{(\alpha x)^2}{2!}+\ldots\right)-\left(\frac{(\beta x)}{1!}+\frac{(\beta x)^2}{2!}+\ldots\right)}{x}$$  $$=\lim_{x\to 0}\left\{\left(\alpha +\frac{\alpha^2 x}{2!}+\ldots\right)-\left(\beta+\frac{\beta^2 x}{2!}+\ldots\right)\right\}$$ $$=\left(\alpha +0\right)-\left(\beta+0\right)$$$$=\color{blue}{\alpha-\beta}$$
