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Finding limit

  • Thread starter l46kok
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  • #1
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Homework Statement


Find the limit of

[itex]\frac{x^3-2x^2-9}{x^2-2x-3}[/itex]

as x->3

Homework Equations





The Attempt at a Solution



You factor the bottom portion and top portion, then it looks something like this

[itex]\frac{x(x^2-2x)-9}{(x-3)(x+1)}[/itex]

I feel like I can go further about eliminating the demonimator but I dont know what
 

Answers and Replies

  • #2
dextercioby
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Is 3 a root of the polynomial x^3- 2x^2 - 9 ? If so, what is then factoring of this polynomial ?
 
  • #3
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(x - 3) is a factor of the numerator. You can use either synthetic division or plain old polynomial division to find the other factor.
 
  • #4
SammyS
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... or notice that

[itex]x^3-2x^2-9 = x^3-3x^2+x^2-9[/itex]

and factor by grouping.
 
  • #5
HallsofIvy
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Homework Statement


Find the limit of

[itex]\frac{x^3-2x^2-9}{x^2-2x-3}[/itex]

as x->3

Homework Equations





The Attempt at a Solution



You factor the bottom portion and top portion, then it looks something like this

[itex]\frac{x(x^2-2x)-9}{(x-3)(x+1)}[/itex]
No, you did NOT factor the numerator. That is not what "factor" means.

I assume you tried first just putting x= 3 into the fraction and found that both numerator and denominator were 0 when x= 3. The fact that the numerator was 0 tells you that it has a factor of x- 3. [itex]x^3- 2x^2- 9= (x- 3)(ax^2+ bx+ c)[/itex]
It shouldn't be hard to see what a, b, and c must be.

I feel like I can go further about eliminating the demonimator but I dont know what
 

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