Question

evaluate lim n->infinity { cos(n) / (n) }

Answer 1

hi

Answer 2

0

Answer 3

answer is 0?

Answer 4

can you please explain?

Answer 5

hi

Answer 6

cos(n) wiggles between 1 and -1 and 1/n kills it.

Answer 7

yes i know that kills 1/n but i don't know how to write down when there's problem says EVALUATE ... LOL

Answer 8

make use of squeeze (sandwich) theorem

Answer 9

Using the Squeeze theorem, you can evaluate \[\lim_{n \rightarrow \infty}\frac{ \cos n }{ n }\] evaluated between -1and 1, since Cosine oscillates between -1 and 1.
Do you know how to apply the squeeze theorem to evaluate the limit?

Answer 10

oh okay

Answer 11

\[-1<\frac{ \cos n }{ n }<1\] So.. you want to evaluate cos by itself, and to do so you you'll have \[\frac{ -1 }{ n }<\frac{ \cos n }{ n } < \frac{ 1 }{ n }\] Now that you have all sides with the same denominator, you can evaluate all three by taking the limit of all 3.\[\lim_{n \rightarrow \infty} (\frac{ -1 }{ n }) <\lim_{n \rightarrow \infty} (\frac{ \cos n }{ n })<\lim_{n \rightarrow \infty} (\frac{ 1 }{ n })\] By properly evaluating this function using the squeeze theorem, you should understand that if the left side goes to 0, and the right side goes to 0, that \[\lim_{n \rightarrow \infty} \frac{ \cos n}{ n } \] will go to 0 as well.

Answer 12

As the limit goes to infinity, cos n, as @experimentX and @abb0t stated, will converge to 0, which isthe behavior of cosine as you take infinitely large numbers.

Answer 13

this is great explanation.. thank you so much i will work on it right now!

Answer 14

Thank you for the medal :)