Question: Need to calculate the following limit without using L'Hopital's Rule:

$$ \lim\_{x \to 0} \frac{5x - e^{2x}+1}{3x +3e^{4x}-3} $$

The problem I'm facing is that no matter what I do I still get expression of the form 
$$ \frac{0}{0} $$
I thought maybe to use
$$ t = e^{2x} $$
But I still can't simplify it enough..

Thank you

Answer: 
$$ \lim\_{x \to 0} \frac{5x - e^{2x}+1}{3x +3e^{4x}-3}=\\
 \lim\_{x \to 0} \frac{5x - (1+2x+\frac{(2x)^2}{2!}+o(x^3))+1}{3x +3(1+4x+\frac{(4x)^2}{2!}+o(x^3))-3} =\\
 \lim\_{x \to 0} \frac{3x-\frac{(2x)^2}{2!}-o(x^3)}{15x +3\frac{(4x)^2}{2!}+o(x^3)} =
\lim\_{x \to 0} \frac{3x}{15x } =\frac{3}{15}$$
**Hint**: We can divide by $x $ to get $$\lim\_{x \to 0} \frac {5-(\frac {e^{2x}-1}{2x}(2))}{3+3 (\frac {e^{4x}-1}{4x}(4))} $$ Can you take it from here?