Solving $\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}}$ How to get from 
$$\lim_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = \lim_{n\to\infty} \frac{\frac{\frac{2}{n}}{\tan{\frac{2}{n}}}\cdot\frac{1}{2n^2}+\frac{\frac{3}{n^2}}{\sin{\frac{3}{n^2}}}\cdot\frac{1}{3}}{\frac{\frac{3}{n}}{\sin{\frac{3}{n}}}\cdot \frac{1}{3n^2}+\frac{\frac{2}{n^2}}{\tan{\frac{2}{n^2}}}\cdot\frac{1}{2}}=...=\frac{2}{3}$$
? It doesn't appear like l'Hôpital's Rule was applied here? 
 A: Here is how I'd do with maybe more of an intuitive approach (all of this can be made more rigorous using taylor expansion for example)
$\sin(x)$ goes like $x$ in the neighborhood of the origin,
using that you have (for example):
$\csc(3/n^3)$ behaves like $n^3/3$ for very large $n$
Similarly, $\cos(x)$ can be approximated by $1$ around the origin thus $\cot(1/x)$ goes like $1/(1/x)$ with very large $x$ (ie. goes like $x$).
using that kind of approach, it is rather straightforward to obtain the solution. Hope this helps.
EDIT: hm, I did it for fun on the side and you might have forgotten something as, if I'm not mistaken, the limit does not converge. (Leading term in the numerator goes like $n^4$ and leading term in the denominator goes like $n^3$, to be more precise the leading term of the expansion at infinity is $\frac23 n$). you might want to check your equation but it doesn't change that you can still use that kind of approach for trigonometric functions.
A: I think you have to check your question because i don't think answer is $\frac{2}{3}$. But i am given a solution of that question which you are write here. 
We know that $\displaystyle \lim_{x\to 0}\frac{x}{\sin x}=1$ and $\displaystyle \lim_{x\to 0}\frac{x}{\tan x}=1$.
$\displaystyle \lim_{n\to\infty}\frac{\cot \frac{2}{n}+n\csc \frac{3}{n^{3}}}{\csc \frac{3}{n}+n\cot \frac{2}{n^{2}}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{1}{\tan \frac{2}{n}}+n\frac{1}{\sin \frac{3}{n^3}}}{\frac{1}{\sin \frac{3}{n}}+n\frac{1}{\tan\frac{2}{n^2}}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{\frac{2}{n}}{\tan \frac{2}{n}}.\frac{n}{2}+n\frac{\frac{3}{n^3}}{\sin \frac{3}{n^3}}.\frac{n^3}{3}}{\frac{\frac{3}{n}}{\sin \frac{3}{n}}.\frac{n}{3}+n\frac{\frac{2}{n^2}}{\tan\frac{2}{n^2}}.\frac{n^2}{2}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{n}{2}+n.\frac{n^3}{3}}{\frac{n}{3}+n.\frac{n^2}{2}}$
$=\displaystyle \lim_{n\to\infty}\frac{3n+2n^{4}}{2n+3n^3}$
$=\displaystyle\lim_{n\to\infty}\frac{\frac{3}{n^3}+2}{\frac{2}{n^3}+\frac{3}{n}}$
$=2.$
A: $$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = 2}$$
There should be a typo in the book. ( the second term in the bottom should have $n^3$ )
$$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^3}}} = \frac{2}{3}}$$
