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Mathematics 20 Online
OpenStudy (anonymous):

Use properties of limits to find the indicated limit. It may be necessary to rewrite an expression before limit properties can be applied. will post pic

OpenStudy (anonymous):

OpenStudy (anonymous):

this is limit

OpenStudy (anonymous):

not sure how to factor out numerator

OpenStudy (anonymous):

x approaches 1

OpenStudy (anonymous):

yes that is the limit but if you use that the denominator becomes 0, undefined so i have to factor out numerator. how do i do that?

OpenStudy (anonymous):

if i get the proper answer ill show hold up lol

OpenStudy (anonymous):

u tried synthetic division

OpenStudy (anonymous):

Yes, synthetic division will factor it nicely.

OpenStudy (anonymous):

or just divide

OpenStudy (anonymous):

use synthetic division with 1 x=1 will come out as factor you will be left with quadratic polynomial (x^2+6x+9) so x^3+5x^2+3x-9=(x-1)(x+3)^2

OpenStudy (anonymous):

When you do the synthetic division, the x-1 terms will cancel out so you get a simple limit of a quadratic equation.

OpenStudy (anonymous):

you know it is \(x^3+5x^2+3x-9)=(x-1)(ax^2+bx+c)\) so it does not take a lot of imagination to find \(a=1,c=9\) all you need really is \(b\)

OpenStudy (anonymous):

so the answer is A? 16? since|dw:1345693944162:dw|

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