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Math Help - Limit of a function

  1. #1
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    Limit of a function

    I've got a problem with the following limit..

    <br />
(x -> 0) lim  [(x^3 - 9x^2 + 27x - 27) / (x^2 - x -6)]<br />
    can someone solve this.. the solution is 0 but i dunno how.. i get 9/2
    Last edited by metlx; January 22nd 2009 at 08:53 AM.
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  2. #2
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    Quote Originally Posted by metlx View Post
    (x->0) lim (x3 - 9x2 + 27x - 27) / (x2 - x - 6)
    (btw, how do you use the math tags properly? so the 'x->0' would go below 'lim' and 'x2' would have a uppercase 2)
    \lim_{x \to 0} (x^3 - 9x^2 + 27x - 27) / (x^2 - x - 6)

    \lim _{x \to 0} (x^3 - 9x^2 + 27x - 27) / (x^2 - x - 6)
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  3. #3
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    Hello, metlx!

    I don't agree with their answer either,
    . . but I bet I know where the error is . . .


    \lim_{x\to0}\frac{x^3 - 9x^2 + 27x - 27}{x^2-x-6}

    As it is written, we can simply substitute x = 0

    . . and we get: . \frac{0-0+0-27}{0-0-7} \:=\:\frac{-27}{-6} \;=\;\frac{9}{2}



    I'm certain the limit is: . {\color{red}x\to3}

    . . \lim_{x\to3}\frac{x^3-9x^2+27x-27}{x^2-x-6} \:=\:\frac{0}{0} . . . an indeterminate form.


    We have: . \frac{x^3-9x^2+27x-27}{x^2-x-6} \:=\:\frac{(x-3)^3}{(x-3)(x+2)}

    . . Since x \neq 3, we can reduce: . \frac{(x-3)^2}{x+2}


    Therefore: . \lim_{x\to3}\frac{(x-3)^2}{x+2} \;=\;\frac{0^2}{5} \;=\;0

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  4. #4
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    thanks!

    I've got another problem with roots..

    \lim_{x\to4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}<br />

    the answer is 4/3.. but how?
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  5. #5
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    never mind. i managed to solve it =)

    \lim_{x\to4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2} = \lim_{x\to4}\frac{(1+2x-9) (\sqrt{x}+2)}{(x-4) (\sqrt{1+2x}+3)}= \lim_{x\to4}\frac{ 4(2x-8)}{6(x-4)} = \lim_{x\to4}\frac{8(x-4)}{6(x-4)} = \frac {4}{3}<br /> <br />
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