anonymous
  • anonymous
Lim x goes to 1 x^2-1/|x-1| anyone can find this limit?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\lim_{x\to 1}\frac{x^2-1}{|x-1|}\]
anonymous
  • anonymous
seems unlikely since this is is a piecewise function it is one thing if \(x>1\) and quite another if \(x<1\)
anonymous
  • anonymous
if \(x>1\) then \(|x-1|=x-1\) when you factor and cancel you get \[x+1\]

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anonymous
  • anonymous
you get something else if \(x<1\) try it and see
anonymous
  • anonymous
I wonder if I should use left and right limit to do this
anonymous
  • anonymous
you have no choice but to do that, since \(|x-1|\) is a piecewise function which changes definition at \(x=1\)
anonymous
  • anonymous
lets take it step by step
anonymous
  • anonymous
if \(x>1\) then \(|x-1|=x-1\) right?
anonymous
  • anonymous
did i lose you there?
anonymous
  • anonymous
im here
anonymous
  • anonymous
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 \(x\to 1^+=1+1=2\)
anonymous
  • anonymous
I see, now the left limit I guess
anonymous
  • anonymous
left limit is different because if \(x<1\) then \(|x-1|=1-x\)
anonymous
  • anonymous
so the limit doesnt exist right
anonymous
  • anonymous
no the two sided limit does not exist
anonymous
  • anonymous
i see
anonymous
  • anonymous
Thank you

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