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Math Help - You guessed it... a limit question.

  1. #1
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    You guessed it... a limit question.

    I'm having a bit of a difficult time understanding a problem that asks:

    Find all values of {a} such that

    \lim_{x\to1}\left(\frac{1}{x-1}-\frac{a}{x^2-1}\right)

    exists and is finite.

    I feel that this should be simple, but I am stuck. The following is about as far I can make sense of it:

    \rightarrow  \lim_{x\to1}\left(\frac{1}{x-1}-\frac{a}{(x-1)(x+1)}\right)

    \rightarrow  \lim_{x\to1}\left(\frac{x+1}{(x-1)(x+1)}-\frac{a}{(x-1)(x+1)}\right)

    \rightarrow  \lim_{x\to1}\left(\frac{x+1-a}{(x-1)(x+1)}\right)

    Thinking about numbers very close to 1, I believe a = 2; however, I am unable to work through it.
    Any help in understanding what I am not seeing would be greatly appreciated.

    Thank you in advance.

    Take care.
    /alan
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  2. #2
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    Re: You guessed it... a limit question.

    a=2 will indeed result in a finite limit in this case. If a = 2, then the numerator of your final limit will just be x-1. Then you can reduce the fraction to just 1/(x+1), and limit x --> 1 will exist by plugging in.
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  3. #3
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    Re: You guessed it... a limit question.

    If we use the direct substitution we get \frac{2-a}{0} now for that to have an existing limit

    We must have 2-a=0 the a=2

    so we have \lim_{x\to 1}\frac{x-1}{(x-1)(x+1)}=\frac{1}{2} so the limit exists

    Now we can prove that this is the only solvation by LH rule

    so in order for the limit to exist we must have the case \frac{0}{0} so we can apply LH rule

    \lim_{x\to 1}\frac{1}{2x}=\frac{1}{2}
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  4. #4
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    Re: You guessed it... a limit question.

    So, is determining the value of a really just evaluating

    x+1-a=0

    \rightarrow (1)+1-a=0

    \rightarrow 2-a=0

    \rightarrow a=2

    and then plugging in 2 for a to determine if the limit exists?

    I apologize if this a dumb question. I want to be sure I fully understand every limit problem I do, and sometimes the solutions seem simpler to reach than expected.
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  5. #5
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    Re: You guessed it... a limit question.

    Yea. If you think about it, at x = 1, the denominator will be 0 and there is nothing you can do about it. So you will have a limit of the form b/0, which is +-infinity, unless b is zero itself, because 0/0 is an indeterminate form. To keep the limit finite, you need to balance the infinitesimal size of the denominator by making the numerator infinitesimal as well, as x --> 1.
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