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anonymous
 5 years ago
f(x)=x^4+2x+1
lim x goes to 1 (f(x)4x)/(x^31)=?
anonymous
 5 years ago
f(x)=x^4+2x+1 lim x goes to 1 (f(x)4x)/(x^31)=?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0replace f(x) with it "value" into the 2nd equation.... x^4 +2x +1 4x // x^3 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the top and bottom differ by 1 degree so you asymptote with be a slant or oblique at y=x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i saw the words corectly in my head, but my fingers are forming a coop....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I still can't find solution prof. :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know, i should simplify equation and plug in 1.But can't simplify.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't place 1 for x into the equation as you might with other problems because in \[f(x)= (x^42x+1)/(x^31)\] placing 1 in for x would result in a zero in the denominator. You have to first get rid of the x^31 in the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know , but i can't get rid of the x^31 denominator, i have tried to simplify but couldn't.Can you show how ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i thought it said as x _. infinty...im going blind i tellya ....BLIND!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x>....and my hand eye brain coordination is getting shot....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x^3  1 is a difference of cubes right? x^3  1^3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x1)(x^2 +x +1) is its factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah ,it equals (x1)(x^2+x+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0facotr the top and see if x1 cancles

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to simplify using the differnce of cubes: \[f(x)= (x^42x+1)/((x1)(x^2+x+1))\] You c\]an then use long division on the numerator (divide the whole top by x1 to get [f(x)= (x^3x^2+x1)/(x^2+x+1)\

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The bottom part glitched up but you get the point.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I get: (x1)(x^3 +x^2+x 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah (x1)(x^3 +x^2+x 1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02/3 as the limit? as x>1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0X^2 would have to be negative because you're dividing by (x1). Once you multiply (x1) by x^3 to get rid of the x^4, you would get x^3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it aint negative....i double checked that ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I made Polynomial long division and it's positive , but thanks Athenian

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01  1 0 0 2 1 0 1 1 1 1  1 1 1 1 0 x^3 +x^2 +x 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, good job Athen....really, that was great work :)
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