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Math Help - when x approaches to -inf in limit

  1. #1
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    when x approaches to -inf in limit

    Hi,
    I'm having problems with few limit questions

    lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

    lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

    lim x-> -inf (sqrt (16x^2 - 5x))/(3x+1)


    i have to solve these problems by divide by highest power in denominator thingy
    but i am really confusing about the sign in denominator.


    what i am thinking is that:

    for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

    for the second one, i think i should leave it as it is because the highest power is 2 ?

    for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?


    am i right? if not, could you point me out where i went wrong ?
    Last edited by haebinpark; February 1st 2010 at 06:55 AM.
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    Hi,
    I'm having problems with few limit questions

    lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

    lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

    lim x-> -inf (sqrt 16x^2 - 5x)/(3x+1)


    i have to solve these problems by divide by highest power in denominator thingy
    but i am really confusing about the sign in denominator.


    what i am thinking is that:

    for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

    for the second one, i think i should leave it as it is because the highest power is 2 ?

    for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?


    am i right? if not, could you point me out where i went wrong ?

    There's a trick you can always try: put t:=-x\Longrightarrow x\rightarrow -\infty \Longleftrightarrow t\rightarrow \infty , so for example, and taking into account that A^3+B^3=(A+B)(A^2-AB+B^2):

    \lim_{x\to -\infty}\frac{x^3-1}{2x^2(x-1)}=\lim_{t\to \infty}\frac{-t^3-1}{2t^2(-t-1)}= \lim_{t\to \infty}\frac{t^3+1}{2t^2(t+1)}=\lim_{t\to \infty}\left(\frac{t^2-t+1}{2t^2}\cdot \frac{t^{-2}}{t^{-2}}\right)=\lim_{t\to \infty}\frac{1-1\slash t+1\slash t^2}{2}=\frac{1}{2} .

    Now you try the other ones by yourself

    Tonio
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  3. #3
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    Quote Originally Posted by haebinpark View Post
    Hi,
    I'm having problems with few limit questions

    lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

    lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

    lim x-> -inf (sqrt 16x^2 - 5x)/(3x+1)


    i have to solve these problems by divide by highest power in denominator thingy
    but i am really confusing about the sign in denominator.


    what i am thinking is that:

    for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

    for the second one, i think i should leave it as it is because the highest power is 2 ?

    for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?


    am i right? if not, could you point me out where i went wrong ?
    \lim_{x \to -\infty}{\frac{x^3 - 1}{2x^2(x-1)}}=\lim_{x \to -\infty}{\frac{1 - \frac{1}{x^3}}{2-\frac{1}{x}}}=\frac{1}{2}
    \lim_{x \to -\infty}{\frac{6x^2 - x +3}{3-x-4x^2}}=\lim_{x \to -\infty}{\frac{6 - \frac{1}{x} +\frac{3}{x^2}}{\frac{3}{x^2}-\frac{1}{x}-4}}=\frac{-3}{2}
    \lim_{x \to -\infty}{\frac{\sqrt {16x^2} - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-4x - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-9}{3+\frac{1}{x}}}=-3
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  4. #4
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    so, none of them need to be change the sign in denominator?
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  5. #5
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    thank you for the help tonio,
    but i don't really understand how it goes like that,
    could you explain again in a little easier way?
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  6. #6
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    common procedure

    Quote Originally Posted by lovemath View Post
    \lim_{x \to -\infty}{\frac{x^3 - 1}{2x^2(x-1)}}=\lim_{x \to -\infty}{\frac{1 - \frac{1}{x^3}}{2-\frac{1}{x}}}=\frac{1}{2}
    \lim_{x \to -\infty}{\frac{6x^2 - x +3}{3-x-4x^2}}=\lim_{x \to -\infty}{\frac{6 - \frac{1}{x} +\frac{3}{x^2}}{\frac{3}{x^2}-\frac{1}{x}-4}}=\frac{-3}{2}
    \lim_{x \to -\infty}{\frac{\sqrt {16x^2} - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-4x - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-9}{3+\frac{1}{x}}}=-3
    as lovemath did most elementary problems involving limits to infinity can be solved by dividing by the highest power of x.
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