chris215
  • chris215
-
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
It is important to know the definition of absolute value. |x|=x if x>0 or if x=0 |x|=-x if x<0 -- x>0 think we are looking to the right of 0 --x<0 think we are looking to the left of 0 So |x-8|=x-8 if x-8>0 or if x-8=0 |x-8|=-(x-8) if x-8<0
freckles
  • freckles
Your question says you are looking to the left of 8 which means you should be thinking of values less than 8
freckles
  • freckles
So which one will you use for |x-8|

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freckles
  • freckles
the choices are A) x-8 B -(x-8)
chris215
  • chris215
a?
freckles
  • freckles
but |x-8|=x-8 if x>8 and |x-8|=-(x-8) if x<8
anonymous
  • anonymous
you could yust mathway.com
freckles
  • freckles
you have x is approaches values to the left of 8 not to the right
freckles
  • freckles
since x is approaching to the left you have |x-8|=-(x-8)
freckles
  • freckles
if you had \[\lim_{x \rightarrow 8^+} \frac{|x-8|}{x-8}=\lim_{x \rightarrow 8^+}\frac{x-8}{x-8}=1\] but instead of that little + sign thingy you have the - sign thingy
freckles
  • freckles
\[\lim_{x \rightarrow 8^-}\frac{|x-8|}{x-8}=...\] and that - sign thing means you want to look to the left of 8 as you are approaching 8 which means you are looking at values then 8 which means you are looking at x<8 and in the definition I wrote |x-8|=-(x-8) if x<8 also if you don't see that x<8 if x-8<0 then try adding 8 on both sides to solve the inequality
freckles
  • freckles
@chris215 are you still confused what to do?
freckles
  • freckles
\[\lim_{x \rightarrow 8^-} \frac{|x-8|}{x-8}=\lim_{x \rightarrow 8^-} \frac{-(x-8)}{x-8}=...\]
chris215
  • chris215
-1?
freckles
  • freckles
yes

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