I have no idea need help
2007-04-09 04:39:42 · 8 answers · asked by malasunas 3 in Science & Mathematics ➔ Mathematics
This limit does not exist. cos(x) is periodic, and it will continue to oscillate between -1 and 1. To have a limit, the value of the function must approach that limit such that f(x) is arbitrarily close to the limit for some value f(p) and all values of f(x) for x greater than p, which never happens here.
2007-04-09 04:41:14 · answer #1 · answered by DavidK93 7 · 6⤊ 0⤋
Cos Infinity
2016-10-04 23:05:30 · answer #2 · answered by seligson 4 · 0⤊ 0⤋
This Site Might Help You.
RE:
what is the limit as x--> infinity of cosx?
I have no idea need help
2015-08-16 20:45:37 · answer #3 · answered by Anonymous · 0⤊ 2⤋
This limit doesn't exist. the cosine function is periodic, keeps oscillating from -1 to 1 and doesn't converge to anything. It's bounded, but doesn't have a limit at infinity.
With exception of the constant function, no periodic function has a limit at infinity.
2007-04-09 04:50:23 · answer #4 · answered by Steiner 7 · 6⤊ 0⤋
I've posted a beautifully typed and illustrated solution at the link below. It looks as it would in a book. Look.
http://www.tomsmath.com/does-the-limit-of-cosine-x-as-x-goes-to-infinity-exist.html
2014-06-14 10:38:45 · answer #5 · answered by ? 3 · 3⤊ 0⤋
No limit. cos x varies in a smooth oscillation between +1 and -1, forever.
2007-04-09 04:42:43 · answer #6 · answered by Philo 7 · 4⤊ 0⤋
As (x) ===> infinity the limit of Cos(x)=0
The question ,IS :
as (x) ---> infinity.(definable)
The question IS NOT
as Cos(x),---> infinity, (not definable),
-1 <= cos =>+1
Just from the look of it, 0 would be the obvious answer, and it is
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Consider a triangle with legs along the x and y axis
The angle between (x) and (y) legs is the Cos(x) .
(y) the leg parallel to the (y) axis (height) stays constant .
Extend (x), ,the leg along the x axis ,
(x) ----> infinity
Cos(x) angle ---> 0 . .
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Cos can be used when describing a circles, waves, triangle , slope of a line, all sorts of things
2007-04-09 05:19:44 · answer #7 · answered by Anonymous · 3⤊ 7⤋
you are wrong. you are taking the cos of the ratio of the two legs.
so if the y axis leg stays at 1 and the x axis leg goes to infinity, you are taking the equation limit x->infinity of cos(1/x).....where limit x->infinity of 1/x=0.......you're example is flawed because the answer would be cos(0)=1
2007-04-12 18:44:49 · answer #8 · answered by Anonymous · 0⤊ 8⤋