Finding the value of this limit $$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}$$
The answer is given to be $1$. Can someone give any hint/s related to this problem?
More importantly, why am I not able to get the answer by simply  using L hopital's rule and by basic substitution of trigonometric and exponential limits?
Any kind of help would be appreciated. I gave the problem simply as an example to reflect the above confusion.
EDIT: Even though tag says so, you are still allowed to use l hospital's rule to help evaluate the limit. 
The question is edited now. It is e^(2x^4) in denominator. 
Again, thanks a lot for taking your time to reply to my doubt. 

You can use taylor series (a lot easier than L'Hopital's):
We have the following limit:
$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)}$$
Using taylor series:
$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)} = \lim_{x\to 0}\frac{\left(x^4-\frac{(x^4)^3}{3!} + \frac{(x^4)^5}{5!} - \ldots\right)-x^4\left(1-\frac{(x^4)^2}{2!} + \frac{(x^4)^4}{4!} - \ldots\right)+x^{20}}{x^4\left(\left(1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + \ldots\right) +-1-2x^4\right)}= \frac{\frac{x^{12}}{12} + \ldots}{2x^5 + \ldots + \frac{256x^{12}}{40320} + \ldots} = 0$$
If you notice, our leading coefficient for the numerator is $x^{12}$ and for the denominator is $x^5$.
I don't believe the answer is $1$. Therefore, the answer is $0$. Comment if you have questions.
