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Math Help - solving a limit with a root? please help!

  1. #1
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    solving a limit with a root? please help!

    I'm solving a limit problem that looks something like this:

    lim as x-> 0 ; 1-(sqrt) 2x(sq)-1 / x-1

    each time I attempt to solve this problem, I end up solving in circles and I'm uncertain what I should do. We've only learned basic things about limits and hence I'm confused as to where to start besides multiplying by the 1+ (sqrt) 2x(sq)-1 / 1+ (sqrt)2x(sq)-1. after I've done that I get something to the effect of 1-2x(sq)-1/ (x-1) (1+(sqrt)2x(sq)-1);; can someone please tell me if my math is flawed or what I should do next?
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    Re: solving a limit with a root? please help!

    Quote Originally Posted by BernieBriggs View Post
    I'm solving a limit problem that looks something like this:
    lim as x-> 0 ; 1-(sqrt) 2x(sq)-1 / x-1
    It the problem is \lim _{x \to 0} \frac{{1 - \sqrt {2x^2  - 1} }}{{x - 1}}, then the limit does not exist, because 0 is not in the domain of the function.
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  3. #3
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    Re: solving a limit with a root? please help!

    rather than try "fancy algebraic manipulation" just plug in x = 0. do you notice something odd about the square root?
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    Re: solving a limit with a root? please help!

    It comes out as sqrt of -1; which i know is imaginary. the problem is that my math teacher insists that you can solve it but I don't even understand how.;; and from what you both said it doesn't exist.

    ****EDITTTT!;*** he changed the problem to x approaches 1; what does this change for this problem
    Last edited by BernieBriggs; September 25th 2012 at 03:51 PM. Reason: a change in the situation of the problem
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    Re: solving a limit with a root? please help!

    Quote Originally Posted by BernieBriggs View Post
    ****EDITTTT!;*** he changed the problem to x approaches 1; what does this change for this problem
    In that case:
    \frac{{1 - \sqrt {2x^2  - 1} }}{{x - 1}}=\frac{-2(x+1)(x-1)}{(x-1)(1 + \sqrt {2x^2  - 1})}
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    Re: solving a limit with a root? please help!

    It changes it crucially! x= 1 is in at least on the boundary of the domain of this function. Of course if you just put x= 1 in the formula you get "0/0" which does not give an answer. But it should tell you that you need to do some factoring. looking at that square root in the numerator suggests to me that you should "rationalize the numerator".
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    Re: solving a limit with a root? please help!

    I see what I did wrong; I was just factoring in correctly. Thank you all so much!

    ps; just to affirm I'm not crazy; x(sq)-2x+4; if you plug in -2; do you get 12? (unrelated problem; but someone tried to explain this equals 10 and I was like ? no...)
    Last edited by BernieBriggs; September 25th 2012 at 04:28 PM.
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    Re: solving a limit with a root? please help!

    alright; so now i've factored correctly but I'm still caught in this horrible paradoxic multiplication of roots. i divided out and then multiplied out the root to the top; so now i have 2(x+1) (1- sqrt 2x(sq)-1)/2-2x(sq).

    what've I done wrong, now? : / I'm still getting zero as a denominator and I'm not sure how to fix it.
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    Re: solving a limit with a root? please help!

    \lim _{x \to 1} \frac{{1 - \sqrt {2x^2  - 1} }}{{x - 1}} \cdot \frac{1 + \sqrt {2x^2  - 1}}{1 + \sqrt {2x^2  - 1}}

    \lim _{x \to 1} \frac{1 - (2x^2  - 1)}{(x-1)(1 + \sqrt {2x^2  - 1})}

    \lim _{x \to 1} \frac{2 - 2x^2}{(x-1)(1 + \sqrt {2x^2  - 1})}

    \lim _{x \to 1} \frac{2(1-x^2)}{(x-1)(1 + \sqrt {2x^2  - 1})}

    \lim _{x \to 1} \frac{2(1-x)(1+x)}{(x-1)(1 + \sqrt {2x^2  - 1})}

    \lim _{x \to 1} \frac{-2(1+x)}{(1 + \sqrt {2x^2  - 1})}

    now perform the direct substitution of x = 1
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    Re: solving a limit with a root? please help!

    thank you so much! You do not believe how much over thinking I put into that...
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