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Math Help - Need help with limit

  1. #1
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    Need help with limit

    Hello, I need help with few questions.


    [Solved]
    Lim x^2/absolute value of x =0/0
    x->0

    I don't know how to deal with the absolute value and factoring x^2

    Its already in its lowest form therefore i substitute in zero and got the answer.
    =========================
    [Solved]
    If f(x) = -25/2x+3
    Find Lim = [f(1+h)-f(1)]/h =0/0
    h->0

    This is what i had done for Q2:

    I substitute in f(x)

    = { {-25/[2(1+h)+3]} + [25/(2(1)+3] }/ h
    = { [-25/(2h+5)] + (5) } / h
    = [ (-25+10h+25)/(2h+5) ] / h
    = [ 10h/2h+5 ]/h
    = help
    ======================
    Find c and n

    lim [(4)(x^9-11)(x^3+8)] / [(c)(x^n-8)(x^2+17)] = 4/4 = 1
    x-> infinity

    =======================


    let f(x) = [(3)(x^9-12)(x^3+3)] and let g(x) = [(c)(x^n-9)(x^2+16)] with c cannot = 0
    Lim
    x-> infinity f(x)/g(x)= infinity

    Implies that n (greater or equal or less than) (?)

    for the last two question i have no idea how to start.
    Thank You
    Last edited by hovermet; September 19th 2011 at 09:07 PM. Reason: Figured out number one myself
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help with limit

    Q2
    If f(x)=\frac{-25}{2x+3} then:
    \lim_{h\to 0} \frac{\frac{-25}{2(1+h)+3}+5}{h}
    =\lim_{h \to 0} \frac{\frac{-25+5(5+2h)}{5+2h}}{h}
    =\lim_{h\to 0} \frac{\frac{10h}{5+2h}}{h}
    =\lim_{h\to 0} \frac{10h}{(5+2h)h}
    =\lim_{h \to 0} \frac{10}{5+2h}=2
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  3. #3
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    Re: Need help with limit

    Quote Originally Posted by Siron View Post
    Q2
    If f(x)=\frac{-25}{2x+3} then:
    \lim_{h\to 0} \frac{\frac{-25}{2(1+h)+3}+5}{h}
    =\lim_{h \to 0} \frac{\frac{-25+5(5+2h)}{5+2h}}{h}
    =\lim_{h\to 0} \frac{\frac{10h}{5+2h}}{h}
    =\lim_{h\to 0} \frac{10h}{(5+2h)h}
    =\lim_{h \to 0} \frac{10}{5+2h}=2
    I c, I multiplied the h to 10h that's y i couldn't get the answer.

    Thanks
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Need help with limit

    You're welcome!

    For question 3, you notice you get the indeterminate form \frac{\infty}{\infty} and the limit is rational so ...?
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  5. #5
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    Re: Need help with limit

    As far as the first one is concerned, are you aware that |x|= x if x\ge 0 and |x|= -x if x< 0? Use those two cases to determine what |x|/x is and what the limit is.
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  6. #6
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    Re: Need help with limit

    Quote Originally Posted by Siron View Post
    You're welcome!

    For question 3, you notice you get the indeterminate form \frac{\infty}{\infty} and the limit is rational so ...?
    For indeterminate form of infinity/infinity of a rational function you divide the numerator and denominator by x^n where n is the highest power present in the denominator.

    Both Q3 and Q4 have x^n in the denominator. There isn't a specific value. I cannot just divide by x^n can I? The question gave the limit. How do i work backwards?
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  7. #7
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    Re: Need help with limit

    Yes, so i do both cases. Therefore (0^2)/(0) = 0/0 = 0
    (0^2)/-(0) = 0/0 = 0
    The Lim is zero.

    If the x is not approaching zero, can there be more than one limit?
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  8. #8
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    Re: Need help with limit

    I managed to solved the rest myself, thank you all.
    Last edited by hovermet; September 20th 2011 at 10:53 PM.
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