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: do a simple trick: $\dfrac{e^{3x}-1}{e^{4x}-1} = \dfrac{e^{3x}-1}{3x}\cdot \dfrac{4x}{e^{4x}-1}\cdot \dfrac{3}{4}$. The first two factors tend to $1$ each, leaving you the last factor being the answer.
You can use  l'Hopital's Rule to get 
$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{3e^{3x}}{4e^{4x} }= \frac{3}{4}.$