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Math Help - Limits involving infinity

  1. #1
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    Limits involving infinity

    Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
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    Quote Originally Posted by TsAmE View Post
    Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
    hi
    \lim_{x\to 1}\left (\frac{x}{(x-1)^2}  \right )=\infty
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    Quote Originally Posted by TsAmE View Post
    Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
    It has no limit, as x approaches 1,
    since the graph approaches infinity.
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    What happened to the 2 in the numerator? And if you sub in x=1 for the denominator it equals 0 which is undefined? By the way how did you post the maths equation with the right symbols? I tried copy and pasting the symbols from MS Word, but they didn't come out
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    Quote Originally Posted by TsAmE View Post
    What happened to the 2 in the numerator? And if you sub in x=1 for the denominator it equals 0 which is undefined? By the way how did you post the maths equation with the right symbols? I tried copy and pasting the symbols from MS Word, but they didn't come out
    \lim_{x \to 1} \frac{2-x}{(x-1)^2} = +\infty.
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    Sigh I can't see the pictures, its says the following: To view links or images in signatures your post count must be 10 or greater. You currently have 3 posts.
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  7. #7
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    look at the graph ...
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    Thanks a lot. So is the only way to solve this problem by looking at the graph, as you cant cancel out the denominator?
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    Quote Originally Posted by TsAmE View Post
    Thanks a lot. So is the only way to solve this problem by looking at the graph, as you cant cancel out the denominator?
    no ... you can also think about it.

    as x \to 1 , the numerator approaches 1

    as x \to 1, the denominator approaches a very small positive value (why will it be positive?)

    as a result ... \frac{a \, fixed \, value}{a \, very \, small \, positive \, value} \to \infty
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    Quote Originally Posted by TsAmE View Post
    Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
    Also, you can write

    \frac{2-x}{(x-1)(x-1)}=\frac{1+(1-x)}{(1-x)(1-x)}

    since (x-1)(x-1)=[-(1-x)][-(1-x)]=(1-x)(1-x)

    leading to

    \frac{1}{(1-x)(1-x)}+\frac{(1-x)}{(1-x)(1-x)}

    =\frac{1}{(1-x)(1-x)}+\frac{1}{1-x}

    As x tends to 1, both of these tend to +\infty
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    But if you sub in x = 1.11111111 for the 1/ 1 - x part, the limit becomes negative
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    Quote Originally Posted by TsAmE View Post
    But if you sub in x = 1.11111111 for the 1/ 1 - x part, the limit becomes negative
    That's very true,
    but the first one will be far more positive, giving an overall positive result.

    If x=1.1

    \left(\frac{1}{-0.1}\right)\left(\frac{1}{-0.1}\right)+\frac{1}{-0.1}=\frac{1}{0.01}-\frac{1}{0.1}=100-10

    I should have said, as x tends to 1, the "sum" of these tends to +\infty with the first fraction dominant.
    Last edited by Archie Meade; April 24th 2010 at 01:13 PM.
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    Oh ok I see makes sense. So if I were to do this question in a test without a calculator, how would I know that I must test values close to the limit? Must I do this if I cant seem to cancel out the denominator?
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