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 2 years ago
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?
 2 years ago
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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shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0lim x>inf for all sinx cosx and tanx are undefined..

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0why you say it'll be 0 or 1 ? o.O

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0because thats what occurs most ina period... those are true fax

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0so you are saying that sin(infinity)= 0 ? would that mean, according to you, sin(infinity + pi/2) =1 ?

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0no 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

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.1Look at a graph of sinx, cosx, or tanx, as x gets larger... what happens to the graph?

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0man i wish shub was here so he could put in on this...

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.1Or look at this animation of simple harmonic motion (makes a sinusoidal wave): http://upload.wikimedia.org/wikipedia/commons/7/74/Simple_harmonic_motion_animation.gif No matter how long you watch that, is the amplitude ever going to change?

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1There 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.

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0its something like .. the limit doesnt exist... but it is assumed it would = 0 for sinx and assumed 1 on cosx

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1No it isn't, never, you're flat out wrong.

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0its what my high school teacher and college professor said...

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1Your high school teacher and college professors are wrong or you misheard them.

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0well if u had to guess a number... what would u guess?

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0it was something like probability led the solution to kinda* exist

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1There'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.

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0thats true... but 0 occurs .... tbh i cant explain it like they did

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1The only time you might have a limit with sine or cosine converging towards something might be something like: \[\lim_{x \rightarrow \infty }\frac{ sinx }{ x }=0\]

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.1^ it's probably something like what kainui posted. Damped oscillation.

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0XD this question always gets every1 involved
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