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Thread: another basic limit

  1. #1
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    another basic limit

    How do I evaluate without using L'Hopital's rule
    lim sqrt(x^2-9)/abs(3-x) as x->3-

    I didn't manage to manipulate the algebric expression. I got to sqrt((x+3)/(x-3)) as x->3- and I don't know what to do from here
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  2. #2
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    Quote Originally Posted by GIPC View Post
    How do I evaluate without using L'Hopital's rule
    lim sqrt(x^2-9)/abs(3-x) as x->3-

    I didn't manage to manipulate the algebric expression. I got to sqrt((x+3)/(x-3)) as x->3- and I don't know what to do from here
    Since you are approaching $\displaystyle 3$ from the left, you need to take notice of what happens with the values of $\displaystyle x < 3$.

    Notice that $\displaystyle |x - 3| = -(x - 3) = 3 - x$ if $\displaystyle x < 3$.


    So $\displaystyle \lim_{x \to 3^{-}}\frac{\sqrt{x^2 - 9}}{|x - 3|} = \lim_{x \to 3^{-}} \frac{\sqrt{x^2 - 9}}{3 - x}$

    $\displaystyle = \lim_{x \to 3^{-}}\frac{x^2 - 9}{(3 - x)\sqrt{x^2 - 9}}$

    $\displaystyle = \lim_{x \to 3^{-}}\frac{-(3 - x)(x + 3)}{(3 - x)\sqrt{x^2 - 9}}$

    $\displaystyle = \lim_{x \to 3^{-}}\frac{-(x + 3)}{\sqrt{x^2 - 9}}$.


    This now tends to $\displaystyle -\frac{6}{0}$ which tends to $\displaystyle -\infty$.
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    and how did you conclude that 6/0 tends to infinity? I can't use L'Hopital's rule in this one, how can I explain this result otherwise?
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    Quote Originally Posted by GIPC View Post
    and how did you conclude that 6/0 tends to infinity? I can't use L'Hopital's rule in this one, how can I explain this result otherwise?
    How many 0's go into 6?

    0 + 0 + 0 + ... = 0

    So infinitely many will go into 6.
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    I understand the intuitive explanation but if I had to submit the homework, How do I show it on a more formal note?
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    Quote Originally Posted by GIPC View Post
    I understand the intuitive explanation but if I had to submit the homework, How do I show it on a more formal note?
    It's well known that $\displaystyle \lim_{0 \to \infty}\frac{a}{x} = \infty$ for $\displaystyle a > 0$.

    Check the graph of $\displaystyle \frac{1}{x}$.

    It's one of the most basic limits and one you will need to remember.
    Last edited by Prove It; Apr 14th 2010 at 02:55 AM.
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    Quote Originally Posted by Prove It View Post
    It's well known that $\displaystyle \lim_{x \to \infty}\frac{a}{x} = \infty$ for $\displaystyle a > 0$.

    Check the graph of $\displaystyle \frac{1}{x}$.

    It's one of the most basic limits and one you will need to remember.
    Typo- that should be $\displaystyle \lim_{x\to 0}\frac{a}{x}= \infty$.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Typo- that should be $\displaystyle \lim_{x\to 0}\frac{a}{x}= \infty$.
    Thanks, will edit now.
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