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Thread: Help Finding Limits!!!

  1. #1
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    Help Finding Limits!!!

    I need help finding the answers to these, I did some of them but are unsure of the answer.

    -Use the limit laws to find the limit if it exists:

    1. lim x=> 0- square root of -x
    i got 0 for this one, but i'm not sure

    2. lim x=> 8 absolute value of (x-8)/x-8
    i'm not entirely sure on this one

    3. lim x=> 0 (1-2x)^3/x
    i got zero on this one

    4. lim x=> infinity 3e^3x+2e^x-1/e^3x-e^2x+5
    no idea on this one


    Any help would be a big help, thanks
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  2. #2
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    Quote Originally Posted by roksteady View Post
    I need help finding the answers to these, I did some of them but are unsure of the answer.

    -Use the limit laws to find the limit if it exists:

    1. lim x=> 0- square root of -x
    i got 0 for this one, but i'm not sure

    2. lim x=> 8 absolute value of (x-8)/x-8
    i'm not entirely sure on this one

    3. lim x=> 0 (1-2x)^3/x
    i got zero on this one

    4. lim x=> infinity 3e^3x+2e^x-1/e^3x-e^2x+5
    no idea on this one


    Any help would be a big help, thanks
    You are correct for 1)

    For 2) you must split it into two limits. The absoulte value of a function, f(x) is $\displaystyle f(x) \iff f(x) \geq 0$ but $\displaystyle -f(x) \iff f(x) < 0 $

    So treat it as two cases:
    The first case is when x approaches 8 through positive values, when this happens f(x) > 0, hence |f(x)| = f(x).

    $\displaystyle \displaystyle \lim_{x \to 8^+} \frac{x-8}{x-8} = \displaystyle \lim_{x \to 8^+} 1 = 1 $

    Now through negative values, when this happens f(x) <0, hence |f(x)| = -f(x).

    $\displaystyle \displaystyle \lim_{x \to 8^-} \frac{8-x}{x-8} = \displaystyle \lim_{x \to 8^-} -1 = -1 $

    Since these limits are not equal, the limit does not exist!

    For 3):

    Use the rules $\displaystyle \displaystyle \lim_{x \to a} (f(x))^n = (\displaystyle \lim_{x \to a} f(x))^n $ and the rules $\displaystyle \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x \to a} f(x)}{\displaystyle \lim_{x \to a} g(x)} $

    This gives:

    $\displaystyle \displaystyle \lim_{x \to 0} \frac{(1-2x)^3}{x} = \frac{(\displaystyle \lim_{x \to 0} (1-2x))^3}{\displaystyle \lim_{x \to 0} (x)} = \frac{1}{0} = \pm \infty$. Limit no existy!


    For 4) Is this the equation you meant? $\displaystyle \displaystyle \lim_{x \to \infty} 3e^{3x}+\frac{2e^{x-1}}{e^{3x}}-e^{2x+5}$

    If so:

    $\displaystyle \displaystyle \lim_{x \to \infty} 3e^{3x}+\frac{2e^{x-1}}{e^{3x}}-e^{2x+5} =\displaystyle \lim_{x \to \infty}(3e^{3x}) + \frac{\displaystyle \lim_{x \to \infty} (2e^{x-1})}{\displaystyle \lim_{x \to \infty} (e^{3x})} -\displaystyle \lim_{x \to \infty} (e^{2x+5}) $

    $\displaystyle \displaystyle =3 \lim_{x \to \infty}(e^{x})^3 + 2 \frac{\lim_{x \to \infty} (e^x e^{-1})}{\lim_{x \to \infty} (e^{x})^3} -\lim_{x \to \infty} (e^{2x}e^5) $

    $\displaystyle \displaystyle =3 \bigg(\lim_{x \to \infty}(e^{x})\bigg)^3 + 2e^{-1} \frac{ \displaystyle \lim_{x \to \infty} (e^x)}{\displaystyle \bigg(\lim_{(x \to \infty} (e^{x})\bigg)^3} -e^5 \displaystyle \lim_{x \to \infty} (e^{x})^2 $

    $\displaystyle \displaystyle =3 \bigg(\lim_{x \to \infty}(e^{x})\bigg)^3 + 2e^{-1} \frac{ \displaystyle \lim_{x \to \infty} (e^x)}{\displaystyle \bigg(\lim_{x \to \infty} (e^{x})\bigg)^3} -e^5\bigg(\displaystyle \lim_{x \to \infty} (e^{x})\bigg)^2 $

    Now the only limit you have to handle is $\displaystyle \displaystyle \lim_{x \to \infty} e^x $.
    Last edited by Mush; Feb 6th 2009 at 06:12 PM.
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