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Math Help - Pre-Calculus Question w/ Some Basic Calculus Involved

  1. #1
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    Pre-Calculus Question w/ Some Basic Calculus Involved

    Hi,

    \lim_{\substack{x-3}}  f(x)=\frac{x^2+4x-21}{x-3}.

    Using L'Hopital's rule (or algebra manipulation) we can determine the limit is 10. My question is, how is this possible if it is a rational function with a vertical asymptote at 3? Shouldn't it approach infinity?

    After graphing it, I noticed that it happens to be a straight line and virtually equal to the graph of its derivative f'(x)=2x+4? This seems odd, as the graph said there was a hole at the point x=3, not a vertical asymptote. If I slightly change the function (such as x-4 in the denominator), it looks more like a rational function with a vertical asymptote.

    I don't need any computational help with this problem - I just want to understand what's going on here.
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  2. #2
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    Nevermind - I realized after factoring the numerator that the factor (x-3) from the denominator cancels out, essentially creating the function y=x+7 with a hole at x=3, equivelent to the entire function when you graph it.

    This is what I get for teaching myself calculus while only just starting pre-calculus.
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