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: 
Using Taylor's series:

$$\frac{5x - e^{2x}+1}{3x +3e^{4x}-3} = \frac{5x - (1+2x + O(x^2))+1}{3x +3(1+4x+O(x^2))-3} = \frac{3x - O(x^2)}{15x+ O(x^2)} .$$

Thus, $$\lim\_{x \rightarrow 0} \frac{3x - O(x^2)}{15x+ O(x^2)} = \frac{1}{5}.$$
Do you know the Taylor formula? If so, then
$$
e^x = 1 + x + o(x)
$$
as $x \to 0$. Here, $o(x)$ is a function such that $o(x)/x \to 0$ as $x \to 0$.
Hence
$$
\frac{5x - e^{2x}+1}{3x +3e^{4x}-3} = \frac{5x - (1 + 2x + o(x)) + 1}{3x + 3(1 + 4x + o(x))-3} = \frac{3x+ o(x)}{15x+o(x)} = \frac{3+o(x)/x}{15+o(x)/x}
$$
which shows that your limit is $1/5$.