anonymous
  • anonymous
lim x approaches 1 |(x-1)|/(x-1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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AravindG
  • AravindG
What did you get when you tried to evaluate this?
anonymous
  • anonymous
0? i plugged in 1 but i dont think thats correct..
AravindG
  • AravindG
See first convert modulus into function at different intervals |x|=x when x>0 |x|=-x when x<0 Write this for |x-1|

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anonymous
  • anonymous
don't be hoodwinked by this
anonymous
  • anonymous
\[\frac{|x-1|}{x-1}\] is only ever one of two numbers if \(x-1>0\) then it is \(1\) and if \(x-1<0\) then it is \(-1\) try a few numbers and you will see it instantly
anonymous
  • anonymous
so i just plug in numbers greater and less than 1?
anonymous
  • anonymous
lets go slow
anonymous
  • anonymous
what i meant was, if you plug in any number larger than 1 for x, you will get 1
anonymous
  • anonymous
\[\frac{|3-1|}{3-1}=\frac{2}{2}=1\] for example
anonymous
  • anonymous
so for any number larger than 1, this expression is just the number 1
anonymous
  • anonymous
and for any number less than one, this expression is just the number \(-1\) for example, if \(x=0\) you get \[\frac{|0-1|}{0-1}=\frac{1}{-1}=-1\]
anonymous
  • anonymous
if x - 1 > and < aren't equal, does that mean the limit doesn't exist?
anonymous
  • anonymous
in other words \[\frac{|x-1|}{x-1} = \left\{\begin{array}{rcc}- 1& \text{if} & x <1 \\ 1& \text{if} & x>1 \end{array} \right. \]
anonymous
  • anonymous
and yes, since the limit from the right is 1 and from the left is -1, that means the limit does not exist, like you said
anonymous
  • anonymous
ok thank you i understand
anonymous
  • anonymous
yw

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