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Math Help - limits pass papers

  1. #1
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    limits pass papers

    \[ \lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]
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  2. #2
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    Re: limits pass papers

    So...
    \ \lim_{x \to \4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]
    You forgot the TEX tags.

    The overall limit is indeterminate, although from the right side, denoted by  \lim_{x \to 4+}, it should be infinity.
    (Note the plus is on the other side of the four)

    The easiest way to do this is to graph it then approximate the value by plugging in an obscure number such as 4.0000000000000001 instead of 4, this should give you some idea.
    Last edited by Xeritas; March 18th 2012 at 04:23 PM.
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  3. #3
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    Re: limits pass papers

    Quote Originally Posted by mathkid12 View Post
    \[ \lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]
    \displaystyle \lim_{x \to 4^{+}} \frac{\sqrt[2]{x}}{x^2 - 5x + 4}=\displaystyle \lim_{x \to 4^{+}} \frac {\sqrt[2]{x}}{(x-4)(x-1)}=\frac{2}{0^{+} \cdot 3} = \frac{2}{0^{+}}=+\infty
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