To find the limit of $\frac{1}{\sin n}+\frac{1}{\cos n}$ What should be the value of $\lim (\frac{1}{\cos n}+\frac{1}{\sin n})$ ? I think the limit does not exist.
Thanks in advance
 A: Let $x_n=1/\sin n+1/\cos n$. 


*

*The jumps from $n$ to $n+1$ are of size $1$ and the size of a quadrant is $\pi/2\gt1$ so there are infinitely many integers $n$ such that $n$ mod $2\pi$ is in each of the four quadrants.

*For every $n$ in the first quadrant $(0,\pi/2)$ mod $2\pi$, one knows that $0\lt\cos n\lt1$ and $0\lt\sin n\lt1$ hence $1/\cos n\gt1$ and $1/\sin n\gt1$ hence $x_n\gt2$.

*For every $n$ in the third quadrant $(\pi,3\pi/2)$ mod $2\pi$, one knows that $-1\lt\cos n\lt0$ and $-1\lt\sin n\lt0$ hence $1/\cos n\lt-1$ and $1/\sin n\lt-1$ hence $x_n\lt-2$.


This proves that $(x_n)$ diverges.
A: Notice that $$\lim (\frac{1}{\cos n}+\frac{1}{\sin n})=\lim \frac{\sin n+\cos n}{\sin n\cos n}=\lim \frac{2\tan \frac n2 - \tan^2 \frac n2 + 1}{2\tan \frac n2}.$$ 
If this limit exists, say $A$. Let $g(n)=\frac{2\tan \frac n2 - \tan^2 \frac n2 + 1}{2\tan \frac n2}$. Clearly $\lim g(n)=A$.
By solving $\tan \frac n2$ in the equation $g(n)=\frac{2\tan \frac n2 - \tan^2 \frac n2 + 1}{2\tan \frac n2}$ and finding that $\lim \tan\frac n2$, however, $\lim \tan \frac n2$ does not exist. A contradiction!
A: (This should be a comment, but I don't have the reputation needed)\
Paul's solution could be completed noticing that, though the existence of $ \lim g(n) $ does not imply the existence of $ \lim tan(\dfrac{n}{2}) $, it does imply that $ {tan(\dfrac{n}{2})} $ can be separated into (at most) two sequences with different limits by the same argument stated by Paul ($tan\frac{n}{2}\neq 0$ for $n\in N$ is a consequence of $\pi$ irrational ).\
The contradiction appears when we take in account (as in the comments) that $ U=\{n+2k\pi , n\in N\:, k\in Z\} $ is dense in $R^+$. (Suppose the existence of a positive minimal member of $ U $ and use the reminder division and the irrationality of $ \pi $ to get a contradiction, and because of the definition of $ U $, if $ u\in U $ then $ ku\in U, k\in N $)
