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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?

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Something is wrong with the expression, if the limit is to be $2/3$. The top blows up far faster than the bottom. – André Nicolas Apr 19 '12 at 7:51
Look at displayed line in your question. On the left you have $n\csc\frac{3}{n^3}$. Then after the first equals sign, in the same position you have in the denominator $\sin \frac{3}{n^2}$. Which is it? Is it $n^3$ or is it $n^2$? If it involved $\frac{3}{n^2}$, limit would exist. But with $\frac{3}{n^3}$, which is in title and another place, limit will not exist. – André Nicolas Apr 20 '12 at 0:59
up vote 1 down vote accepted

$$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = 2}$$

There should be a typo in the book. ( the second term in the bottom should have $n^3$ )

$$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^3}}} = \frac{2}{3}}$$

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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.

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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 $\displaystyle \lim_{x\to 0}\frac{x}{\tan x}=1$.

$\displaystyle \lim_{n\to\infty}\frac{\cot \frac{2}{n}+n\csc \frac{3}{n^{3}}}{\csc \frac{3}{n}+n\cot \frac{2}{n^{2}}}$ $=\displaystyle \lim_{n\to\infty}\frac{\frac{1}{\tan \frac{2}{n}}+n\frac{1}{\sin \frac{3}{n^3}}}{\frac{1}{\sin \frac{3}{n}}+n\frac{1}{\tan\frac{2}{n^2}}}$ $=\displaystyle \lim_{n\to\infty}\frac{\frac{\frac{2}{n}}{\tan \frac{2}{n}}.\frac{n}{2}+n\frac{\frac{3}{n^3}}{\sin \frac{3}{n^3}}.\frac{n^3}{3}}{\frac{\frac{3}{n}}{\sin \frac{3}{n}}.\frac{n}{3}+n\frac{\frac{2}{n^2}}{\tan\frac{2}{n^2}}.\frac{n^2}{2}}$ $=\displaystyle \lim_{n\to\infty}\frac{\frac{n}{2}+n.\frac{n^3}{3}}{\frac{n}{3}+n.\frac{n^2}{2}}$ $=\displaystyle \lim_{n\to\infty}\frac{3n+2n^{4}}{2n+3n^3}$ $=\displaystyle\lim_{n\to\infty}\frac{\frac{3}{n^3}+2}{\frac{2}{n^3}+\frac{3}{n}}$ $=2.$

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$\displaystyle \lim_{n\to\infty}\frac{\cot \frac{2}{n}+n\csc \frac{3}{n^{3}}}{\csc \frac{3}{n}+n\cot \frac{2}{n^{3}}}=\frac{2}{3}.$ I think this is a right question. – Kns Apr 19 '12 at 15:39

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