Calculate $\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$ Question:
Calculate 
$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$$
using substitution, cancellation, factoring etc. and common standard limits (i.e. not by L'Hôpital's rule).
Attempted solution:
It is not possible to solve by evaluating it directly, since it leads to a "$\frac{0}{0}$" situation. These kinds of problems are typically solved by cancellation and/or direct application of a after some artful substitution in a precalculus context.
It was a bit hard to find a good substitution, so I tried several:
$t = 3x$
$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=3x\} = \lim_{x \to 0} \frac{e^{t} - 1}{e^{4\frac{t}{3}} - 1}$$
Although this gets me closer to the standard limit
$$\lim_{x to 0} \frac{e^x - 1}{x} = 1$$
...it is not good enough.
$t = e^{4x} - 1$
$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=e^{4x} - 1\} = \lim_{t \to 0} \frac{e^{\frac{3 \ln (t+ 1)}{4}} - 1}{t}$$
Not sure where to move on from here, if it is at all possible.
Looks like the denominator can be factored by the conjugate rule:
$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \lim_{x \to 0} \frac{e^{3x} - 1}{(e^{2x} - 1)(e^{2x} + 1)}$$
Unclear where this trail can lead.
What are some productive substitutions or approaches to calculating this limit (without L'Hôpital's rule)?

Hint:  You presumably already know how to prove L'hospital's Rule in general.  Try adapting that general proof to this specific case.
