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Math Help - Limits

  1. #1
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    Post Limits

    Hi everyone!

    I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

    Question 1.
    Evaluate the limit of \frac{(x+1)}{(|x|-1)} as x\rightarrow-1

    Answer:
    If x=-1 then get \frac{-1+1}{1-1}=\frac{0}{0}!

    Therefore let x=-1-\in then \in\rightarrow0 in limit (\in >0)

    i.e. \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1

    x=-1-\in=-(1+\in)
    |-1-\in|=(1+\in)

    If let x=-1+\in=-(1-\in) and |-1+\in|=1-\in (\in>0)

    Therefore
    \frac{x+1}{|x|-1} with x=-1+\in is \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1 same answer as before

    Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in x=-1-\in. Hopefully someone can help.

    Thank you

    Dadon
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dadon View Post
    Hi everyone!

    I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

    Question 1.
    Evaluate the limit of \frac{(x+1)}{(|x|-1)} as x\rightarrow-1

    Answer:
    If x=-1 then get \frac{-1+1}{1-1}=\frac{0}{0}!

    Therefore let x=-1-\in then \in\rightarrow0 in limit (\in >0)

    i.e. \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1

    x=-1-\in=-(1+\in)
    |-1-\in|=(1+\in)

    If let x=-1+\in=-(1-\in) and |-1+\in|=1-\in (\in>0)

    Therefore
    \frac{x+1}{|x|-1} with x=-1+\in is \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1 same answer as before

    Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in x=-1-\in. Hopefully someone can help.

    Thank you

    Dadon
    You use x=1\pm \epsilon, \epsilon > 0 as you want to look at the behaviour of the expression near x=-1, and you need to handle the \pm \epsilon cases seperatly to control the behaviour of |1 \pm \epsilon| while looking at the limit from above and below.

    RonL
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  3. #3
    Junior Member
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    Jan 2007
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    Quote Originally Posted by dadon View Post
    Hi everyone!

    I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

    Question 1.
    Evaluate the limit of \frac{(x+1)}{(|x|-1)} as x\rightarrow-1

    Answer:
    If x=-1 then get \frac{-1+1}{1-1}=\frac{0}{0}!

    Therefore let x=-1-\in then \in\rightarrow0 in limit (\in >0)

    i.e. \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1

    x=-1-\in=-(1+\in)
    |-1-\in|=(1+\in)

    If let x=-1+\in=-(1-\in) and |-1+\in|=1-\in (\in>0)

    Therefore
    \frac{x+1}{|x|-1} with x=-1+\in is \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1 same answer as before

    Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in x=-1-\in. Hopefully someone can help.

    Thank you

    Dadon
    similar to previous explanation

    if h > 0

    (a + h) with h-->0 means you approach the point 'a' from the right of 'a' (on the number line)

    (a - h) with h-->0 means you approach the point 'a' from the left of 'a' (on the number line)

    the corresponding limiting values of the function are the right hand and left hand limits respectively

    the function has a limit as x-->a only if the left hand and right hand limits are equal.
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