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Thread: One sided Limits

  1. #1
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    One sided Limits

    I was posed this problem on another forum by a first year and for some reason I'm struggling

    lim x-> 1 from the left of
    $\displaystyle (1-2x)/(x^2-1)$

    Could someone provide me a precise way to show the limit goes to infinity?
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  2. #2
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    I use a theorem : let $\displaystyle {x_k}->1$ from left as k->infinite , and f be continuous if $\displaystyle {f(x_k)}->infinite$ , $\displaystyle limf(x)=infinite$as $\displaystyle x->1^-$

    clearly , $\displaystyle (1-2x)/(x^2-1)$ is a continuous on $\displaystyle (-infinite , 1^-)$ and let $\displaystyle {x_k}={(k-1)/k}$converges to 1 from left.

    And $\displaystyle f(x_k)={(k^2)/2}-k/(4k-2)$ this is infinite as k->infinite
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  3. #3
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    Quote Originally Posted by chipai View Post
    I use a theorem : let $\displaystyle {x_k}->1$ from left as k->infinite , and f be continuous if $\displaystyle {f(x_k)}->infinite$ , $\displaystyle limf(x)=infinite$as $\displaystyle x->1^-$

    clearly , $\displaystyle (1-2x)/(x^2-1)$ is a continuous on $\displaystyle (-infinite , 1^-)$ and let $\displaystyle {x_k}={(k-1)/k}$converges to 1 from left.

    And $\displaystyle f(x_k)={(k^2)/2}-k/(4k-2)$ this is infinite as k->infinite
    Oh yes! I have heard a theorem like that before. So I follow you up until the point where you let $\displaystyle {x_k}={(k-1)/k}$. It is an increasing function, but is there a way to find such {x_k}? EDIT: or is it arbitrary?
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  4. #4
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    Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
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  5. #5
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    Quote Originally Posted by chipai View Post
    Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
    thank you for your help.
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  6. #6
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    think simple: $\displaystyle x^2 - 1$ approaches 0 from negative values because $\displaystyle x \rightarrow 1-,$ and $\displaystyle 1 - 2x$ approaches $\displaystyle -1.$ so the limit is $\displaystyle +\infty.$
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    think simple: $\displaystyle x^2 - 1$ approaches 0 from negative values because $\displaystyle x \rightarrow 1-,$ and $\displaystyle 1 - 2x$ approaches $\displaystyle -1.$ so the limit is $\displaystyle +\infty.$
    Well, I was afraid that such an answer was much too simple. :P
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