Standard limits not working on this particular question

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

Answer 1

Use L'Hôpital's rule.

Let $n(x) = 2x +x \cos x -3 \sin x$, $d(x) = x^4\sin x$.

Note that $n^{(k)}(0) = d^{(k)}(0) = 0$ for $k=1,2,3,4$.

However $n^{(5)}(0) = 2, d^{(5)}(0) = 120$.

Answer 2

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 $\boxed{\frac{1}{60}}$.