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Math Help - Another limit question.

  1. #1
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    Another limit question.

    I got another question that was lim as x->3 (3x-8)/(4x-12). I didn't see if it was possible to factor anything so I found that the limit doesn't exist because the left hand limit doesn't equal the right hand limit. for this question I just wanted to know if you could have found another answer algebraically.
    Last edited by mr fantastic; July 10th 2010 at 01:56 AM. Reason: Moved from another thread.
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    Quote Originally Posted by darksoulzero View Post
    I got another question that was lim as x->3 (3x-8)/(4x-12). I didn't see if it was possible to factor anything so I found that the limit doesn't exist because the left hand limit doesn't equal the right hand limit. for this question I just wanted to know if you could have found another answer algebraically.
    \lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)

     \to \infty.
    Last edited by mr fantastic; July 10th 2010 at 01:56 AM.
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    Quote Originally Posted by Prove It View Post
    \lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)

     \to \infty.
    I'm not really sure but if the limit goes to infinity does it exist? I'm not sure what your answer means.
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    Quote Originally Posted by Prove It View Post
    \lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}

     = \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)

     \to \infty.
    Since \displaystyle {\lim_{x \to 3^+}\frac{3x - 8}{4x - 12} = + \infty} and \displaystyle {\lim_{x \to 3^-}\frac{3x - 8}{4x - 12} = - \infty} the limit does not exist.

    Note that f(x) = 1 + \frac{1}{3x - 9} is a hyperbola. It has a vertical asymptote at x = 3 and the above limits are exactly what you see when you draw its graph.
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    I think I've made a mistake here...

    If the function approaches 3 from the left, then we have the denominator with negative values, so approaching -\infty, while if the function approaches 3 from the right, then we have the denominator with positive values, so approaching \infty.

    Therefore the function does not exist.
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    Thanks for your help; I was able to find the answer by evaluating the left and right limit, but I wasn't sure how to do it algebraically.
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    "Trying" to set x= 3 and observing that you get " \frac{1}{0}" is sufficient to conclude that the limit does not exist. Of course, saying that the limit is \infty is just another way of saying there is no limit.
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    Quote Originally Posted by HallsofIvy View Post
    "Trying" to set x= 3 and observing that you get " \frac{1}{0}" is sufficient to conclude that the limit does not exist. Of course, saying that the limit is \infty is just another way of saying there is no limit.
    I would argue there's significant difference between something like \displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty} on the one hand and \displaystyle{ \lim_{x \to 1} \frac{1}{x-1}} which does not exist (and in particular, is not equal to \infty) on the other hand.
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    Quote Originally Posted by mr fantastic View Post
    I would argue there's significant difference between something like \displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty} on the one hand and \displaystyle{ \lim_{x \to 1} \frac{1}{x-1}} which does not exist (and in particular, is not equal to \infty) on the other hand.
    I agree, it depends on the context in which the symbol \infty is being used.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    I would argue there's significant difference between something like \displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty} on the one hand and \displaystyle{ \lim_{x \to 1} \frac{1}{x-1}} which does not exist (and in particular, is not equal to \infty) on the other hand.
    Yes, it is saying that the limit does not exist, for a specific reason and so is more precise than just saying "the limit does not exist". But it is still saying "the limit does not exist".
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