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kinda silly but lim y=sin(x) x-> infinity this would be 0 since that occurs most in a given period and lim y=cos(x) x->inifity is 1 because that occurs most ina period... but what would be limit y=tan(x) x-> infinity ??? would it be undefined because you can take each individual limit and get 1/0= undefined?

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lim x->inf for all sinx cosx and tanx are undefined..
why you say it'll be 0 or 1 ? o.O
because thats what occurs most ina period... those are true fax

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Other answers:

ofcorse not..
i meant to @Goten77
so you are saying that sin(infinity)= 0 ? would that mean, according to you, sin(infinity + pi/2) =1 ?
no shub because like with any limits if i had like 1/(x+1) the limit as x-> infiity is not really impacted by the +1 so in ur example the pi/2 doenst really effect infiity
Look at a graph of sinx, cosx, or tanx, as x gets larger... what happens to the graph?
man i wish shub was here so he could put in on this...
Or look at this animation of simple harmonic motion (makes a sinusoidal wave): No matter how long you watch that, is the amplitude ever going to change?
There is no limit as x goes to infinity of sinx or cosx, this is a fact. They are constantly oscillating between -1 and 1, so you can't say for certain that it becomes anything at infinity.
its something like .. the limit doesnt exist... but it is assumed it would = 0 for sinx and assumed 1 on cosx
No it isn't, never, you're flat out wrong.
its what my high school teacher and college professor said...
Your high school teacher and college professors are wrong or you misheard them.
well if u had to guess a number... what would u guess?
it was something like probability led the solution to kinda* exist
There's no guessing involved, as you increase to infinity sine and cosine functions do not converge towards anything. They will keep going between -1 and +1 forever.
thats true... but 0 occurs .... tbh i cant explain it like they did
The only time you might have a limit with sine or cosine converging towards something might be something like: \[\lim_{x \rightarrow \infty }\frac{ sinx }{ x }=0\]
^ it's probably something like what kainui posted. Damped oscillation.
XD this question always gets every1 involved

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