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Math Help - One sided limit

  1. #1
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    One sided limit

    Find the value of the one sided limit
    \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}

    WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?


    Thank you very much for all your help!
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  2. #2
    Senior Member TriKri's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the value of the one sided limit
    \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}

    WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?


    Thank you very much for all your help!
    Since both the numerator and the denominator \longrightarrow 0, use líHopitalís rule.
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  3. #3
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    I got it now

    i got it now
    since the denominator is undefined
    \lim_{x \to 2^-} \frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}
    etc...


    - Adam
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  4. #4
    Senior Member TriKri's Avatar
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    \frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}\ =

    \sqrt {(2-x)(1+x)}\ \to \left/\lim_{x \to 0}\right/ \to

    \sqrt {0\cdot (1+x)}\ =\ 0
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the value of the one sided limit
    \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}

    WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?
    Factor, your idea was correct.

    \lim_{x\to 2} \frac{(2-x)(2+x)}{\sqrt{(2-x)(1+x)}}

    Now when we approach the limit from the right,
    x>2 that means,
    2-x<0
    1+x>0
    But then,
    (2-x)(1+x)<0
    And the function is undefined.
    (Radical is a negative number!)

    So the limit (the regular limit) does not exist.
    However, the one-handed limit, might. We already shown that the right-handed does not exists. Let us see the left-handed.

    When we approach the limit from the left,
    x<2 on some open interval (\delta, 2) where \delta <2 (a really small number). Because we do not care what happens far away one what happens really close to the point.
    In that case,
    2-x>0
    1+x>0
    Thus, the radical is defined.
    And furthermore we can factor,
    \lim_{x\to 2^-} \frac{(2-x)(2+x)}{\sqrt{2-x}\cdot \sqrt{1+x}}
    \lim_{x\to 2^-} \frac{\sqrt{2-x}(2+x)}{\sqrt{1+x}}=\frac{0\cdot 4}{3}=0
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    Since both the numerator and the denominator \longrightarrow 0, use líHopitalís rule.
    But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

    -Dan
    Yes you can on the half interval.

    In fact, the proof of the L'Hopital theorem is done in two parts. Left and Right hand limits.
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  8. #8
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    just because the denominator tends to zero does not mean that you can use l'hopital.

    for instance

    \lim_{x\to 1}\frac{1}{1-x}

    it is obvious that the limit does not exist.
    in order to use l'hopital u need to have an indeterminate form.
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  9. #9
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    Quote Originally Posted by putnam120 View Post
    just because the denominator tends to zero does not mean that you can use l'hopital.

    for instance

    \lim_{x\to 1}\frac{1}{1-x}

    it is obvious that the limit does not exist.
    in order to use l'hopital u need to have an indeterminate form.
    I think he was asking whether you can use it here. In this example. Because the numerator is not continous on an open interval conting 1, rather on a half-open interval.
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  10. #10
    Senior Member TriKri's Avatar
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    Quote Originally Posted by topsquark View Post
    But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

    -Dan
    Well, I don't know, but I think so You just have to choose whether x > 2 or x < 2 I guess ... do you have an example of when it doesn't work?
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    Well, I don't know, but I think so You just have to choose whether x > 2 or x < 2 I guess ... do you have an example of when it doesn't work?
    I don't have an example of where it doesn't work, I was simply speculating. I was just trying to be careful since the derivative of \sqrt{2 + x - x^2} doesn't exist at x = 2 (because the function doesn't exist for x < 2.) Apparently as long as the limit of the derivative exists, L'Hopital is usable, if I'm understanding TPH correctly.

    -Dan
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  12. #12
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    Quote Originally Posted by topsquark View Post
    Apparently as long as the limit of the derivative exists, L'Hopital is usable, if I'm understanding TPH correctly.
    As long as the "left handed" or "right handed" derivates exist.
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  13. #13
    Senior Member TriKri's Avatar
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    But why doesn't the right hand exist in this case? Isn't complex numbers allowed?
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  14. #14
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    Quote Originally Posted by TriKri View Post
    Isn't complex numbers allowed?
    First I would like to mention square roots of of negative numbers are undefined. Note, it is mathematically incorrect to say \sqrt{-4}=2i (I know that is what they teach in high schools, but it is still wrong). I have already discussed that on the forum. Next, all the L'Hopitals rule was only proven for real differenciable functions (at least I think). Finally this problem it self is real, implies that the function the poster was using mapped the real numbers into the real numbers.
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  15. #15
    Senior Member TriKri's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    First I would like to mention square roots of of negative numbers are undefined. Note, it is mathematically incorrect to say \sqrt{-4}=2i (I know that is what they teach in high schools, but it is still wrong).
    My world picture has been shattered! ;_;
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