$$ \underset{x\to 0}{\lim} \left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$$
What I've tried so far:
$2+\cos x=1+(1+\cos x)$
$=1+\sin^2\dfrac{x}{2}$
But, I do not think this step is fruitful as I am getting stuck thereafter. Kindly provide some sort of help or hint. Thanks in advance!
Try with Taylor series.
$$\sin(x) \approx x - \frac{x^3}{6}$$
$$\cos(x) \approx 1 - \frac{x^2}{2} + \frac{x^4}{24}$$
The result of the limit will be
$$\frac{1}{60}$$