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Finding a Limit


Date: 02/28/2002 at 22:47:04
From: Corrie
Subject: Limits

My calculus teacher gave my class the answer to this limit; however, 
we cannot figure out how to get the answer that is given. If you could 
help me with a proof, I would be very grateful. 

        1 - sqrt(2x^2 - 1)
    lim ------------------
    x->1      x - 1

Thanks!


Date: 03/01/2002 at 08:47:05
From: Doctor Peterson
Subject: Re: Limits

Hi, Corrie.

If we just replace x with 1, we get (1-1)/(1-1), which we can't 
evaluate; so we'd like to rearrange the expression so we can cancel 
something out. The big thing that is in our way is the "1-sqrt" in the 
numerator; how about if we "rationalize the numerator" by multiplying 
by the conjugate? Then (for x not equal to 1):

    1 - sqrt(2x^2 - 1)   1 + sqrt(2x^2 - 1)
    ------------------ * ------------------ =
          x - 1          1 + sqrt(2x^2 - 1)

          1 - (2x^2 - 1)                 2(1 - x^2)
    --------------------------- = ---------------------------
    (x - 1)(1 + sqrt(2x^2 - 1))   (x - 1)(1 + sqrt(2x^2 - 1))

Can you see the next step? Factor the numerator, cancel, and you will 
have an expression that can be evaluated at x = 0, and which is equal 
to the original expression for all other x, so that its value at x = 0 
is the limit you want.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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