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Math Help - Evaluating Limits

  1. #1
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    Evaluating Limits

    Hi, I'm new here and this is my first post but I'm having trouble with one of my homework problems. I have the function:

    g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.

    a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

    b) Evaluate the limit of g(x) as x->c.

    c) Does the limit g(x) as x->∞ exist?

    I'm completely stuck and unsure where to start...
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  2. #2
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    Re: Evaluating Limits

    Quote Originally Posted by daveMathews48 View Post
    g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.
    a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

    b) Evaluate the limit of g(x) as x->c.

    c) Does the limit g(x) as x->∞ exist?

    I'm completely stuck and unsure where to start...
    −∞−∞
    This is very good question to teach you about limits.

    The answer to a) is f~\&~b. Now you reply giving the reasons.
    Then maybe we can do more.
    Thanks from topsquark
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    Re: Evaluating Limits

    If I cancel out the (x-c) from both the top and bottom then set the new denominator e(x-f)(x-b) to 0? Wouldn't that leave me with e, f, and b?
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  4. #4
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    Re: Evaluating Limits

    Quote Originally Posted by daveMathews48 View Post
    If I cancel out the (x-c) from both the top and bottom then set the new denominator e(x-f)(x-b) to 0? Wouldn't that leave me with e, f, and b?
    No it does not. Because e is a positive number.
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    Re: Evaluating Limits

    Ah that's right, thanks. So for part b with (x-c) cancelled out, how do I evaluate that limit?
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    Re: Evaluating Limits

    For part b) the answer is a fraction. The numerator of which is a(c+b)(c+d).
    What is its denominator?
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    Re: Evaluating Limits

    Are you substituting c for x? That would make the denominator e(c-f)(c-b)?
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    Re: Evaluating Limits

    Also, I was thinking about part c. Doesn't lim g(x) as x-> ∞ look for values where x is really large? If I expand the numerator and denominator I get:

    (a(x^2+bx+dx+bd)) / (e(x^2-bx-fx+bf)), which for large x leaves me with ax^2/ex^2? Does that mean the limit exists and is a/e?
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    Re: Evaluating Limits

    Quote Originally Posted by daveMathews48 View Post
    Are you substituting c for x? That would make the denominator e(c-f)(c-b)?
    In limits NEVER JUST SUBSTITUTE!. Ask what happens near the limiting value?
    If x is near c then (x-b) is near c-b.

    Now for the last part. We play the 'back-of-the-book" game.
    You look up the answer. You see: "the limit exists and equals \frac{a}{e}".
    You may be scratching your head why? Can you explain why?
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  10. #10
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    Re: Evaluating Limits

    Ah ok, that makes sense. Well, I don't know if that's the right answer this is one of those problems that's not in the back of the book :P. I mean I understand I can reduce the fraction that way because when x is really large, x^2 is much more important than anything else in the equation. And when I have x^2/x^2 that just equals 1.
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