# Thread: lim [x->2^+] [8/(x^2 - 4) - x/(x - 2)]: tried multiplying common denom's, then....

1. ## lim [x->2^+] [8/(x^2 - 4) - x/(x - 2)]: tried multiplying common denom's, then....

I evaluated the limit by multiplying common denominators then distributed terms so I have polynomials on top and bottom. Then derived each separately and got -6x/6x=-1 but the limit is actually -3/2. What's up with that?

. . . . .$\displaystyle \lim_{x \rightarrow 2^+}\, \left(\dfrac{8}{x^2\, -\, 4}\, -\, \dfrac{x}{x\, -\, 2}\right)$

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2. Originally Posted by Seed5813
I evaluated the limit by multiplying common denominators then distributed terms so I have polynomials on top and bottom. Then derived each separately and got -6x/6x=-1 but the limit is actually -3/2. What's up with that?

. . . . .$\displaystyle \lim_{x \rightarrow 2^+}\, \left(\dfrac{8}{x^2\, -\, 4}\, -\, \dfrac{x}{x\, -\, 2}\right)$
Could you show how you got to -6x/6x?

3. Originally Posted by lev888
Could you show how you got to -6x/6x?
(8/(x^2-4))-(x/(x-2))
(8(x-2)-x(x^2-4))/(x^2-4)(x-2)
(8x-16-x^3+4x)/(x^3-6x+8)
8x+4x=12x
Lay hospital's rule:
(12-3x^2)/(3x^2-6)
Again:
(-6x/6x)
-1

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4. 1) Why would you EVER see an x^3 on this problem? Why not seek a LEAST common denominator? Please factor first and see if you can simplify this process.

2) Why would you use l'hospital's rule on this problem. Its just an algebra problem. if you do not see why, try more factoring.

5. Originally Posted by Seed5813
(8/(x^2-4))-(x/(x-2))
(8(x-2)-x(x^2-4))/(x^2-4)(x-2)
(8x-16-x^3+4x)/(x^3-6x+8)
8x+4x=12x
Lay hospital's rule:
(12-3x^2)/(3x^2-6)
Again:
(-6x/6x)
-1

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Everything you did is fine, up to this point:

$\dfrac{8(x-2)-x(x^2-4)}{(x^2-4)(x-2)}$

In the next step, you made an error when expanding the denominator. This is what you should have gotten:

$\dfrac{8x-16-x^{3}+4x}{x^{3}-2x^{2}-4x+8}$

You appear to have dropped the squared part of the term, mistaking it for 2x. With this correction made, everything should work out correctly. Try proceeding from here and see what you get.

6. Originally Posted by tkhunny
1) Why would you EVER see ab x^3 on this problem? Why not seek a LEAST common denominator? Please factor first and see if you can simplify this process.

2) Why would you use l'hospital's rule on this problem. Its just an algebra problem. if you do not see why, try more factoring.
1 polynomials are easy to derive
2 I'm on a chapter about l'hopital's rule

Take a nap

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7. Originally Posted by ksdhart2
Everything you did is fine, up to this point:

$\dfrac{8(x-2)-x(x^2-4)}{(x^2-4)(x-2)}$

In the next step, you made an error when expanding the denominator. This is what you should have gotten:

$\dfrac{8x-16-x^{3}+4x}{x^{3}-2x^{2}-4x+8}$

You appear to have dropped the squared part of the term, mistaking it for 2x. With this correction made, everything should work out correctly. Try proceeding from here and see what you get.
Thank you

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8. Originally Posted by Seed5813
(8/(x^2-4))-(x/(x-2))
(8(x-2)-x(x^2-4))/(x^2-4)(x-2)
(8x-16-x^3+4x)/(x^3-6x+8)
8x+4x=12x
Lay hospital's rule:
(12-3x^2)/(3x^2-6)
Again:
(-6x/6x)
-1

Sent from my LGLS755 using Tapatalk
(x^2-4)(x-2) does not equal (x^3-6x+8). That is where your mistake is. Also, getting that common denominator is NOT the best one to use. Try using the least common denominator.

9. Originally Posted by Seed5813
1 polynomials are easy to derive
2 I'm on a chapter about l'hopital's rule

Take a nap

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1) Polynomials are where you made your errors. You may wish to rethink your position on this.
2) Your chapter on l'hospital does NOT say, "Use this wherever you can and never use anything else." If there is a simple, algebraic solution, you should do that. Does your chapter on l'hospital tell you that it doesn't ALWAYS work? It should.

I suggest you try to learn, rather than trying to rely on tricks. The latter is not going to help you.

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