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anonymous

  • one year ago

Lim x goes to 1 x^2-1/|x-1| anyone can find this limit?

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  1. anonymous
    • one year ago
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    \[\lim_{x\to 1}\frac{x^2-1}{|x-1|}\]

  2. anonymous
    • one year ago
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    seems unlikely since this is is a piecewise function it is one thing if \(x>1\) and quite another if \(x<1\)

  3. anonymous
    • one year ago
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    if \(x>1\) then \(|x-1|=x-1\) when you factor and cancel you get \[x+1\]

  4. anonymous
    • one year ago
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    you get something else if \(x<1\) try it and see

  5. anonymous
    • one year ago
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    I wonder if I should use left and right limit to do this

  6. anonymous
    • one year ago
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    you have no choice but to do that, since \(|x-1|\) is a piecewise function which changes definition at \(x=1\)

  7. anonymous
    • one year ago
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    lets take it step by step

  8. anonymous
    • one year ago
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    if \(x>1\) then \(|x-1|=x-1\) right?

  9. anonymous
    • one year ago
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    did i lose you there?

  10. anonymous
    • one year ago
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    im here

  11. anonymous
    • one year ago
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    if \(x>1\) then \(|x-1|=x-1\)so your function is \[\frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\] making the limit as \(x\to 1^+=1+1=2\)

  12. anonymous
    • one year ago
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    I see, now the left limit I guess

  13. anonymous
    • one year ago
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    left limit is different because if \(x<1\) then \(|x-1|=1-x\)

  14. anonymous
    • one year ago
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    so the limit doesnt exist right

  15. anonymous
    • one year ago
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    no the two sided limit does not exist

  16. anonymous
    • one year ago
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    i see

  17. anonymous
    • one year ago
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    Thank you

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