<s>[INST] 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
 [/INST] 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 $\boxed{\frac{1}{60}}$.
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