Standard limits not working on this particular question I have this question and its been troubling me for so long. I try to use the standard trig. limits but that just fails everytime and I get the answer as $\infty$.
$$\lim_{x \to 0} \frac{2x+x\cos(x)-3\sin(x)}{x^4\sin(x)}$$
note: I have posted the question by already taking the lcm since i didnt think it would matter (hopefully).
I even checked out some limit calculators but all they show is l hopital rule which is very tiring, but they end up with the right answer which is $1/60$. 
More than the answer im trying to figure out why standard limits fail here?
Thank you in advance

One way you can find this is by using the Taylor series expansions for $\cos x$ and $\sin x$.
Using the first few terms in each,
\begin{align*}
\frac{2x + x \cos x - 3 \sin x}{x^4 \cdot \sin x} &= \frac{2x + x(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)) - 3(x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7))}{x^4(x + O(x^3))}\\
&= \frac{2x + x - \frac{x^3}{2} + \frac{x^5}{24} + O(x^7) - 3x + \frac{x^3}{2} - \frac{x^5}{40} + O(x^7)}{x^5 + O(x^7)}\\
&= \frac{\frac{x^5}{60} + O(x^7)}{x^5 + O(x^7)}\\
&= \frac{1}{60} + O(x^2)
\end{align*}
As $x \to 0$, we get that the limit is $\boxed{\frac{1}{60}}$.
