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)?
 A: 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.
A: Here's another way.
$$\frac{e^{3x}-1}{e^{4x}-1}=\frac{(e^x)^3-1}{(e^x)^4-1}=\frac{(e^x-1)(e^{2x}+e^x+1)}{(e^{2x}+1)(e^x+1)(e^x-1)}=\frac{e^{2x}+e^x+1}{(e^{2x}+1)(e^x+1)}.$$
So,
$$\lim_{x\to0}\frac{e^{3x}-1}{e^{4x}-1}=\lim_{x\to0}\frac{e^{2x}+e^x+1}{(e^{2x}+1)(e^x+1)}=\frac{1+1+1}{2\cdot2}=\frac34.$$
A: Use the identities 
$a^3 - b^3 = (a - b) (a^2 + a b + b^2)$
and $a^4 - b^4 = (a - b) (a+b)(a^2 +  b^2)$ where $a=e^x$ and $b=1$ to get
$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{(e^{x} - 1)(e^{2x} +e^{x}+ 1)}{(e^{x} - 1)(e^{x} + 1)(e^{2x} + 1)}= \lim_{x \to 0} \frac{(e^{2x} +e^{x}+ 1)}{(e^{x} + 1)(e^{2x} + 1)}=\frac{3}{4}$
A: Hint: $$\frac{e^{3x}-1}{e^{4x}-1} = \frac{e^{3x}-1}{3x}\cdot\frac{4x}{e^{4x}-1}\cdot \frac{3}{4}$$
A: $$\frac{(e^x)^3-1}{(e^x)^4-1}=\frac{e^x+1}{2(e^{2x}+1)}+\frac{1}{2(e^x+1)}$$
A: Hint:  You presumably already know how to prove L'hospital's Rule in general.  Try adapting that general proof to this specific case.
A: For positive integers
$m$ and $n$,
$\begin{array}\\
\lim_{x \to 0} \frac{e^{mx}-1}{e^{nx}-1}
&=\lim_{x \to 0} \frac{(e^x-1)(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}{(e^x-1)(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}\\
&=\lim_{x \to 0} \frac{e^{(m-1)x}+e^{(m-2)x}+...+e^x+1}{e^{(m-1)x}+e^{(m-2)x}+...+e^x+1}
\quad\text{(true for  $x \ne 0)$}\\
&= \frac{\lim_{x \to 0}(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}{\lim_{x \to 0}(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}
\quad\text{(since the numerator and denominator are $> 1)$}\\
&=\frac{m}{n}
\end{array}
$
Then put
$m=3$ and $n=4$.
A: You can use  l'Hopital's Rule to get 
$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{3e^{3x}}{4e^{4x} }= \frac{3}{4}.$
