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

  1. #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?

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

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    Last edited by stapel; 03-07-2018 at 05:35 PM. Reason: Typing out the text in the graphic; creating useful subject line.

  2. #2
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    Quote Originally Posted by Seed5813 View Post
    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?

    . . . . .[tex]\displaystyle \lim_{x \rightarrow 2^+}\, \left(\dfrac{8}{x^2\, -\, 4}\, -\, \dfrac{x}{x\, -\, 2}\right)[/tex]
    Could you show how you got to -6x/6x?
    Last edited by stapel; 03-07-2018 at 05:36 PM. Reason: Copying typed-out graphical content into reply.

  3. #3
    Quote Originally Posted by lev888 View Post
    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. #4
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    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.
    Last edited by tkhunny; 03-05-2018 at 10:26 PM.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
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    Quote Originally Posted by Seed5813 View Post
    (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
    Everything you did is fine, up to this point:

    [tex]\dfrac{8(x-2)-x(x^2-4)}{(x^2-4)(x-2)}[/tex]

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

    [tex]\dfrac{8x-16-x^{3}+4x}{x^{3}-2x^{2}-4x+8}[/tex]

    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. #6
    Quote Originally Posted by tkhunny View Post
    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. #7
    Quote Originally Posted by ksdhart2 View Post
    Everything you did is fine, up to this point:

    [tex]\dfrac{8(x-2)-x(x^2-4)}{(x^2-4)(x-2)}[/tex]

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

    [tex]\dfrac{8x-16-x^{3}+4x}{x^{3}-2x^{2}-4x+8}[/tex]

    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. #8
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    Quote Originally Posted by Seed5813 View Post
    (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.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  9. #9
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    Quote Originally Posted by Seed5813 View Post
    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.
    Last edited by stapel; 03-07-2018 at 05:37 PM. Reason: Tweaking tone.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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