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: 
HINT:

As $x\to0,x\ne0$ so safely divide numerator & denominator by $x$

and use 

$$\lim\_{h\to0}\dfrac{e^h-1}h=1$$

Observe that the exponent of $e,$ the limit variable and the denominator are **same**.
$$ \frac{5x - e^{2x}+1}{3x +3e^{4x}-3}=\frac12\frac{\dfrac52-\dfrac{e^{2x}-1}{2x}}{\dfrac34+3\dfrac{e^{4x}-1}{4x}}\to\frac12\frac{\dfrac52-L}{\dfrac34+3L}.$$ (by a scaling of $x$, the two ratios tend to $L$).

Then you can take for granted that $L=1$, giving the answer $\dfrac15$.