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. 
 A: If you can't yet use Taylor expansions, you can still use l'Hôpital with some simplifications to begin with.
Edited question
For the question where $e^{2x^4}$ is in the denominator instead of $e^{2x}$, the first step to do is the substitution $t=\sqrt[4]{x}$, noting that the function is even; so surely
$$
\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)
$$
as soon as one of them exists. The limit becomes
$$
\lim_{t\to0^+}\frac{\sin t-t\cos t+t^5}{t(e^{2t}-1-2t)}
$$
We can look for $k$ such that
$$
\lim_{t\to0^+}\frac{e^{2t}-1-2t}{t^k}
$$
exists, finite and nonzero: with l'Hôpital
$$
=\lim_{t\to0^+}\frac{2e^{2t}-2}{kt^{k-1}}
=\lim_{t\to0^+}\frac{4e^{2t}}{k(k-1)t^{k-2}}
$$
which is $2$ as soon as $k=2$. So the original limit can be written
$$
\lim_{t\to0^+}\frac{\sin t-t\cos t+t^5}{2t^3}
\frac{2t^2}{e^{2t}-1-2t}
$$
and the second factor can be disregarded, so the limit to compute is
$$
\lim_{t\to0^+}\left(\frac{\sin t-t\cos t}{2t^3}+t^2\right)
$$
and the second summand can be disregarded. With l'Hôpital on the first summand,
$$
\lim_{t\to0^+}\frac{\sin t-t\cos t}{2t^3}=
\lim_{t\to0^+}\frac{\cos t-\cos t+t\sin t}{6t^2}=
\lim_{t\to0^+}\frac{\sin t}{6t}=\frac{1}{6}
$$
With Taylor expansions available,
$$
\lim_{t\to0^+}\frac{\sin t-t\cos t+t^5}{t(e^{2t}-1-2t)}=
\lim_{t\to0^+}\frac{\sin t-t\cos t+t^5}{t(1+2t+4t^2/2!+o(t^3)-1-2t)}=
\lim_{t\to0^+}\frac{\sin t-t\cos t+t^5}{2t^3+o(t^4)}
$$
so we need to stop Taylor expansion at the $t^3$ terms in the numerator:
$$
\sin t-t\cos t+t^5=t-\frac{t^3}{3!}+o(t^4)-
t\left(1-\frac{t^2}{2!}+o(t^4)\right)+t^5=
\frac{1}{3}t^3+o(t^4)
$$
and so we have
$$
\lim_{t\to0^+}\frac{\frac{1}{3}t^3+o(t^4)}{2t^3+o(t^4)}=\frac{1}{6}
$$
Original question
Let's look if we can find $k$ such that
$$
\lim_{x\to0}\frac{e^{2x}-1-2x^4}{x^k}
$$
is finite and non zero. We can proceed supposing that we have an indeterminate form $0/0$ until the derivative of the denominator becomes constant, so we have 
$$
\lim_{x\to0}\frac{e^{2x}-1-2x^4}{x^k}=
\lim_{x\to0}\frac{2e^{2x}-8x^3}{kx^{k-1}}
$$
which is non zero as soon as $k=1$; so we conclude that
$$
\lim_{x\to0}\frac{e^{2x}-1-2x^4}{x}=2
$$
and we can transform the original limit into
$$
\lim_{x\to 0}
\frac{\sin x^4-x^4\cos x^4+x^{20}}{2x^5}
\frac{2x}{(e^{2x}-1-2x^4)}
$$
The second factor can then be disregarded, because its limit is $1$. So we're left with
$$
\lim_{x\to 0}
\left(\frac{\sin x^4-x^4\cos x^4}{2x^5}+\frac{x^{15}}{2}\right)
$$
and the second summand has limit $0$, so we can disregard it. Now we have something more manageable:
$$
\lim_{x\to 0}
\frac{\sin x^4-x^4\cos x^4}{2x^5}
$$
where we can apply the substitution $t=\sqrt[4]{x}$ (when $x>0$), so let first compute the limit from the right:
$$
\lim_{x\to 0^+}\frac{\sin x^4-x^4\cos x^4}{2x^5}=
\lim_{t\to 0^+}\frac{\sin t-t\cos t}{2t^{5/4}}\overset{\mathrm{(H)}}{=}
\lim_{t\to 0^+}\frac{\cos t-\cos t+t\sin t}{(5/2)t^{1/4}}=0
$$
For the limit from the left apply the substitution $t=\sqrt[4]{-x}$, and the result is $0$ as well.
So you can end up by saying
$$
\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)}=0
$$
A: Divide numerator and denominator by $x^4$, giving you
$$\lim_{x\to 0}\frac{\frac{\sin x^4}{x^4}-\cos x^4+x^{16}}{e^{2x^4}-1-2x^4}$$
Now expand the sine, cosine, and exponential into power series (Maclaurin series) with as many terms as needed.
A: 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.
