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Mathematics 20 Online
OpenStudy (anonymous):

find the limit x->infinity of arctan(x-x^4)

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

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

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?

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

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

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

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

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

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

OpenStudy (anonymous):

night

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