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Math Help - finding a limit algebraiclly

  1. #1
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    finding a limit algebraiclly

    I need help finding the limit of the following. lim ( x - 1 - 1 )
    x--> 1 2x-2 X^2

    I am having trouble with the algebra. The limit of a rational function is found by substitution. But in the case of an indeterminate as this one I must factor, rationalize, combine rational expressions, etc... to simplify into an equivalent function. Then find the limit of the equivalent function by substitution. Can someone please help. I have been trying to figure this out for about six hours. Is there a method that I am unaware of for finding the common denominator and subtracting these three expressions so I can sunstitute and get the limit. My calculator soves the problem but I need to know how to do it by hand.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I need help finding the limit of the following. lim ( x - 1 - 1 )
    x--> 1 2x-2 X^2
    it is not at all clear what function you are taking the limit of

    could you resubmit to clarify this?
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  3. #3
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    solving a limit algebraically

    The exercise in my text says to find the limit algebraically. the problem is: lim as x approaches 1 then the quantity (x/2x-2 minus 1/x^2 minus 1). The answer in the back of the book is 3/4. The answer I get on my ti-89 titanium is -7/2. Hope this clarifies. Thanks for your time.
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  4. #4
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    Quote Originally Posted by bosmith View Post
    I need help finding the limit of the following. lim ( x - 1 - 1 )
    x--> 1 2x-2 X^2
    I think you mean \lim_{x\to 1}\frac{x}{2x-1}- \frac{1}{x^2} but that's easy: the numerator on the first fraction goes to 1 as does the denominator so its limit is 1. The second fraction has numerator 1 while the denominator goes to 1 so its limit is also 1. The limit of the difference is 1- 1= 0.

    I am having trouble with the algebra. The limit of a rational function is found by substitution. But in the case of an indeterminate as this one I must factor, rationalize, combine rational expressions, etc... to simplify into an equivalent function. Then find the limit of the equivalent function by substitution. Can someone please help. I have been trying to figure this out for about six hours. Is there a method that I am unaware of for finding the common denominator and subtracting these three expressions so I can sunstitute and get the limit. My calculator soves the problem but I need to know how to do it by hand.
    Three expressions? Is it \lim_{x\to 1}\frac{x}{2x}- \frac{1}{2x^2}- 1 or img.top {vertical-align:15%;} ?<br /> <br />
  If it is the former, then \frac{x}{2x}- \frac{1}{2x^2}= \frac{x^2- 1}{2x^2}" alt="\lim_{x\to 1}\frac{x}{2x}- \frac{1}{2}- \frac{1}{x^2} ?<br /> <br />
  If it is the former, then \frac{x}{2x}- \frac{1}{2x^2}= \frac{x^2- 1}{2x^2}" /> which, at x= 1, is 0 so the limit of the entire thing is 1. If the latter, then [tex]\frac{x}{2x}- \frac{1}{2}- \frac{1}{x^2}= \frac{x^2- x^2- 2}{x^2}= \frac{2}{x^2}[/itex] and that is 2 at x= 1.

    I can see no "indeterminate" form in any of these. If you can't use LaTex, please use parentheses to make clear exactly what numerators and denominators the fractions have.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Start with

    x/(2x-2) - 1/(x^2-1)

    = x/[2(x-1)] - 1/(x-1)(x+1)

    Use a common denominator of 2(x-1)(x+1) to obtain

    [x(x+1)- 2]/[2(x-1)(x+1)]

    =(x^2 +x -2)/[2(x-1)(x+1)]

    = ([(x-1)(x+2)]/[2(x-1)(x+1)]

    = (x+2)/2(x+1)

    Now taking the limit as x -> 1 = 3/4
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  6. #6
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    Quote Originally Posted by bosmith View Post
    The exercise in my text says to find the limit algebraically. the problem is: lim as x approaches 1 then the quantity (x/2x-2 minus 1/x^2 minus 1). The answer in the back of the book is 3/4. The answer I get on my ti-89 titanium is -7/2. Hope this clarifies. Thanks for your time.
    Ah! It is \lim_{x\to 1}\frac{x}{2x-2}- \frac{1}{x^2- 1}.
    It would have been a lot clearer if you had written x/(2x-2)- 1/(x^2- 1). Use parentheses!

    \frac{x}{2x-2}- \frac{1}{x^2- 1}= \frac{x}{2(x-1)}- \frac{1}{(x-1)(x+1)} has "common denominator" 2(x-1)(x+1) and is \frac{x(x+1)- 2}{2(x-1)(x+1)}= \frac{x^2+ x- 2}{2(x-1)(x+1)} = \frac{(x-1)(x+2)}{2(x-1)(x+1)}. As long as x is not 1, that is the same as \frac{x+2}{2(x+1)} and has the same limit at 1: \frac{1+2}{2(1+1)}= \frac{3}{4}.
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  7. #7
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    HALLSOFIVY, yes, the limit is the three as you described it. I will learn how to use the latex but until then I will use the parenthesis as you stated. the limit as x appraoches 1 is the three rationals ((x/2x-2) - (1/x^2) - 1). All three inside a parenthesis.
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