Mathematics
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OpenStudy (anonymous):
find the limit x->infinity of arctan(x-x^4)
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OpenStudy (anonymous):
this is indeterminate, right?
OpenStudy (anonymous):
let y = lim arctan ( x-x^4) , so take tan of y
tan (y) = tan ( lim arctan (x-x^4)) = lim (tan (arctan ( x-x^4)) = lim x-x^4
OpenStudy (anonymous):
i would go like this:
lim arctan(x( 1- x^3))
OpenStudy (anonymous):
im from waterloo too
OpenStudy (anonymous):
then we have arctan(-infinity)=-pi/2
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OpenStudy (anonymous):
this make sense?
OpenStudy (anonymous):
hmmmm i see got it thx,
OpenStudy (anonymous):
k np
OpenStudy (anonymous):
what about b, im having trouble eliminating the sart sign
OpenStudy (anonymous):
this is a great site- i find the smartest ppl are on between 10pm and midnight, and prob like 9am-noon
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OpenStudy (anonymous):
mwgc, isnt that indeterminate though?
OpenStudy (anonymous):
i went and took 9x^6 out of the sqrt so then we have
3x^3(sqrt(1-x/9x^6))
OpenStudy (anonymous):
mg, no this problem
l
imit x->infinity of arctan(x-x^4)
OpenStudy (anonymous):
then factor out x^3 outta the bottom
OpenStudy (anonymous):
what?
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OpenStudy (anonymous):
he/she is talkin now about part b of this question which is on our assignment
OpenStudy (anonymous):
i dont see part b)
can we talk about part a) please
OpenStudy (anonymous):
i dont really see how you got 3x^3(sqrt(1-x/9x^6))
OpenStudy (anonymous):
arctan ( x - x^4) is indeterminate , arctan (-oo - oo ). oh wait
OpenStudy (anonymous):
sqrt(9x^6) is 3x^3
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OpenStudy (anonymous):
, it isnt indeterminate, it is arctan -oo
OpenStudy (anonymous):
OMG LMAO OMFG
OpenStudy (anonymous):
then when we multiply it, its multiplied by sqrt(1- x/9x^6)
OpenStudy (anonymous):
we divide the second term to cancel out that multiplication
OpenStudy (anonymous):
ya @barry so -pi/2
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OpenStudy (anonymous):
oo - oo is indeterminate, but
-oo - oo is NOT indeterminate
OpenStudy (anonymous):
yo @mwgc, what is part b), i dont see it bro
OpenStudy (anonymous):
ill post it
OpenStudy (anonymous):
lim x->infinity
(sqrt(9x^6-x))/(x^3+1)
OpenStudy (anonymous):
its negative infinity
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OpenStudy (anonymous):
how did u get that i got 3
OpenStudy (anonymous):
lim sqrt ( 9x^6 -x ) / ( x^3 + 1) = sqrt ([9x^6] ( 1 - 1/(9x^5) ) / ( x^3+1) = 3 |x^3| sqrt ( 1 - 1/(9x^5))/(x^3 + 1)
OpenStudy (anonymous):
3x^3sqrt(1- x/9x^6)/x^3(1-1/x^3)
3(sqrt(1-1/9x^5)/1-1/x^3
plug in infinity for x
OpenStudy (anonymous):
right barry is right, use absolute values for takin outta sqrt
OpenStudy (anonymous):
in this case tho doesnt matter
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OpenStudy (anonymous):
. well... it kinda does, but it doesnt effect the solution no
OpenStudy (anonymous):
cuz |positive infinity| is positive infinity
OpenStudy (anonymous):
i thought he wanted negative infinity. sorry
OpenStudy (anonymous):
o crap! sry veryconfused i thot you were saying the answer was -infinity, youre right we are lookin at x->-infinity
OpenStudy (anonymous):
doesnt the answer become -3 then
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OpenStudy (anonymous):
veryconfused u there?
OpenStudy (anonymous):
ya im workin out the question right now and i did get -3
OpenStudy (anonymous):
k good i already handed mine in so i said positive 3 o well
OpenStudy (anonymous):
lol
OpenStudy (anonymous):
k man gnite
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OpenStudy (anonymous):
night