# 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.

• Related: math.stackexchange.com/a/991685
– Did
Oct 27, 2014 at 7:45
• I'm sorry- the limit as n approaches what? Oct 27, 2014 at 23:14
• @daOnlyBG $n$ approaches to infinity.
– KON3
Oct 28, 2014 at 8:12

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.

• Presumably n is an integer, hence it cannot be a multiple of π/4.
– Did
Oct 29, 2014 at 9:04
• Oh I did not know tht but in tht case we can take n to be integers in 1st quad and 3rd quad they will be exact opposite Oct 29, 2014 at 10:38
• There are arguments leading to the conclusion, no question about that, the point is that the one in your answer does not, at present (and "exact opposites" in your comment is dubious).
– Did
Oct 29, 2014 at 10:46
• @did Sorry I meant opposite sign, I have made the required edits Oct 29, 2014 at 11:14
• We are definitely closer to a full proof now but there is still another minor problem: there exists some converging sequences with infinitely many positive terms and infinitely many negative terms hence one needs a liiittle more than what you say.
– Did
Oct 29, 2014 at 12:00

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!

• Why would it be easier to know that tan(n) has no limit rather than cos(n) or sin(n) has no limit?
– Did
Oct 27, 2014 at 8:56
• @Did Please notice that it is the limit of $\frac{1}{\sin n}+\frac{1}{\cos n}$.
– Paul
Oct 27, 2014 at 9:02
• Yes (I can read, thanks), and?
– Did
Oct 27, 2014 at 9:48
• It is obvios that the limit of $\sin n$ or $\cos n$ does not exist, however, can we say that the original limit does not exist quickly also?
– Paul
Oct 27, 2014 at 9:51
• It would not--and this is not the point of my first comment. You reduce the exercise to showing that tan(n/2) has no limit (that is, omitting a proof that cos(n) and sin(n) are always nonzero, which you need, and with a mistake in the reduction since tan(n/2)-1/tan(n/2) might have a limit without tan(n/2) having a limit). Hence, why not prove this as well? Actually it seems that the context of your answer is that the sequence n mod 2pi is dense, but then no proof at all is needed...
– Did
Oct 27, 2014 at 10:00

(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 $ku\in U, k\in N$)