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Math Help - limit as x goes to infinity

  1. #1
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    limit as x goes to infinity

    Consider f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
    Take steps to find the limit as x -> infinity and –infinity
    a) Find the limit of f(x) as x goes to infinity
    rewrite the expression sqrt(9x**2 + 6) by extracting x**2 as follows:
    sqrt(9x**2 + 6) = sqrt(x**2 ) sqrt(9x**2 + 6) = Ax sqrt(p(x))
    p(x) = ________________ and A = ____ (constant)
    b) It follows that f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8) = (A’ sqrt(p(x)) / (-6-8/x)
    find p(x) and A’
    c) what is the limit?
    d) Now find as x goes to – infinity (exact same information, but with – infinity.)


    my work has gotten some of these (in class we didn't use the same format webwork expects..)


    f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
    i took the sqrt(x**2 ) out and got (-4 sqrt(9+ 6/x**2) / (-6-8/x)
    For the constant part of filling it in, -4 doesn't work for "A" and neither does - 12 (thought mauybe -4 * sqrt9)
    anyway i found the limit as x goes to infinity which is 2.

    considering i don't know what they want for the value of A, which is a constant, i don't know what they want for A'
    I tired using teh value 0, (derivative of a constant) but that is wrong!

    hope somoene can help me out on this A thing they're wanting of me!!!
    thankyou
    Brittany
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  2. #2
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    Quote Originally Posted by williamb View Post
    Consider f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
    Take steps to find the limit as x -> infinity and –infinity
    a) Find the limit of f(x) as x goes to infinity
    rewrite the expression sqrt(9x**2 + 6) by extracting x**2 as follows:
    sqrt(9x**2 + 6) = sqrt(x**2 ) sqrt(9x**2 + 6) = Ax sqrt(p(x))
    p(x) = ________________ and A = ____ (constant)
    b) It follows that f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8) = (A’ sqrt(p(x)) / (-6-8/x)
    find p(x) and A’
    c) what is the limit?
    d) Now find as x goes to – infinity (exact same information, but with – infinity.)


    my work has gotten some of these (in class we didn't use the same format webwork expects..)


    f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
    i took the sqrt(x**2 ) out and got (-4 sqrt(9+ 6/x**2) / (-6-8/x)
    For the constant part of filling it in, -4 doesn't work for "A" and neither does - 12 (thought mauybe -4 * sqrt9)
    anyway i found the limit as x goes to infinity which is 2.

    considering i don't know what they want for the value of A, which is a constant, i don't know what they want for A'
    I tired using teh value 0, (derivative of a constant) but that is wrong!

    hope somoene can help me out on this A thing they're wanting of me!!!
    thankyou
    Brittany
    It's easiest to get the answer they want by answering the question they ask .....

    They ask you to write \sqrt{9x^2 + 6} in the form Ax \sqrt{p(x)}. So A = 1 and p(x) = 9 + \frac{6}{x^2}.

    So f(x) = \frac{-4 \sqrt{9x^2 + 6}}{-6x - 8} = \frac{-4 \sqrt{9 + \frac{6}{x^2}}}{-6 - \frac{8}{x}}. Compare this with \frac{A' \sqrt{p(x)}}{-6 - \frac{8}{x}} and it's clear that A' = -4.
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