Question

limit n tends to infinity (cos n)/ n

Answer 1

n only takes integer values, right?

Answer 2

whatever be the value of n, note that -1<= cos(x) <=1

Answer 3

cos(n)*

Answer 4

Someone's jittery :D
If you're going to edit comments, I'd have just edited the first one :D

Not sure how to prove that it tends to 0, though. Try epsilon-N proof.

Answer 5

yes limit n goes to infinity

Answer 6

Considering that cos(n) is bounded (like @shubhamsrg said), you can rewrite the limit as
\[\lim_{n\to\infty}\frac{\cos(n)}{n}\le\lim_{n\to\infty}\frac{|\cos(n)|}{n}\le\lim_{n\to\infty}\frac{1}{n}=0.\]
It then follows that
\[\lim_{n\to\infty}\frac{\cos(n)}{n}=0,\]
since
\[n\to\infty.\]

Answer 7

yes, what you have said is correct, there is a theorem if (a_n) is bounded and (b_n) converges to 0 then (a_nb_n) goes to zero
thanks and regards by Senthil

Answer 8

@SithsAndGiggles you are missing something ... the limit could be negative.

Answer 9

@experimentX, I had the squeeze theorem at the back of my mind at the time, I just didn't know how to articulate it.
@senthilkiitm, I meant to say something more along the lines of the following: cos(n) is bounded, so
\[-1\le\cos(n)\le1\]
Dividing both sides by n, you have
\[-\frac{1}{n}\le\frac{\cos(n)}{n}\le\frac{1}{n}\]
The limits of the left- and right-most sides as n approaches infinity are both 0, so you have
\[0\le\lim_{n\to\infty}\frac{\cos(n)}{n}\le0,\]
which tells you that
\[\lim_{n\to\infty}\frac{\cos(n)}{n}=0.\]

Answer 10

all right .. that's okay ... also you could have used,
\[ \lim_{n\to\infty} \left| \frac{\cos(n)}{n} \right | \le\lim_{n\to\infty}\frac{|\cos(n)|}{n}\le\lim_{n\to\infty}\frac{1}{n}=0.\]
They are equivalent.