Question

computing the limit of the series

\[\lim_{n \rightarrow \infty} \frac{n^{3} + \sin(-2011n)}{n^{3} + 110111}\]
= DNE

is this correct?

Answer 1

My reasoning is
\[\lim_{n \rightarrow \infty} \frac{\sin(-2011n)}{n^{3}+110111} = DNE\]

Answer 2

Thus the limit cannot be computed because sin(infinity) = 1,0

Answer 3

Hint:\[|\sin x|\leqslant1\text{, }\forall x\in\mathbb{R}\]

Answer 4

@across makes a good point. This looks like a job for L'Hospital.

Answer 5

It's actually pretty straightforward.

Answer 6

I thought you couldn't compute the limit as x goes to infinity of trig functions

Answer 7

well at least for sin(x) as it goes to 1 and 0

Answer 8

thus a limit is impossible to compute

Answer 9

Never mind: you edited the OP.

Answer 10

yeah well it would be 1 + DNE = DNE

Answer 11

In your new case, break the fraction into two parts and evaluate each individually.

Answer 12

I did

Answer 13

I still don't understand your reasoning behind the DNE.

Answer 14

We don't need to use L'hospital :)
\[\lim_{n \rightarrow \infty} \frac{n^{3} + \sin(-2011n)}{n^{3} + 110111}\]
divide numerator and denominator by n^3
\[\huge\lim_{n \rightarrow \infty} \frac{1 + \frac{\sin(-2011n)}{n^3}}{1 + \frac{110111}
{n^3}}\]
we know
\[|\sin x|\le 1\]
so
a finite quantity by an infinite quantity is zero so
\[\huge\lim_{n \rightarrow \infty}\frac{1+0}{1+0}\]
hence the limits exist

Answer 15

@ash2326 illustrated my point exactly.

Answer 16

My reasoning behind it is that sin(x) has two possible numbers it can be 1 and 0 thus the limit is not determinate.

Answer 17

as it approaches infinity

Answer 18

No, sin can be anything in the real domain \([-1,1]\).

Answer 19

oh sorry you are correct, brain fart

http://www.wolframalpha.com/input/?i=lim+n-%3Einfinity+sin%28n%29

Answer 20

What you should understand here is that you're basically dividing a number between \(-1\) and \(1\) by infinity.

Answer 21

still how can you determine a limit if there is no one number

Answer 22

@Australopithecus we know that it's between -1, 1
so we know it's a finite quantity.
Hence
\[\lim_{n\to \infty} \frac{\sin n}{n}=0\]
if we had
\[\lim_{n\to \infty} {\sin n}\]
then it'd not exist

Answer 23

oh alright so sin(n)/n as n goes to infinity = 0

Answer 24

thanks for clarifying that for me

Answer 25

so the answer is 1

Answer 26

or 1 + 0