$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}$$
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.
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 $$ \lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)}=0 $$