Babynini
  • Babynini
Limits of absolute values
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Babynini
  • Babynini
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Jhannybean
  • Jhannybean
@Astrophysics
Astrophysics
  • Astrophysics
Ok lets first note, |dw:1443917561077:dw| if we plug the limit we will get 0/0 correct?

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Babynini
  • Babynini
Right
Astrophysics
  • Astrophysics
All we really need to know is |dw:1443917619092:dw| so we have to break it up as such
Astrophysics
  • Astrophysics
Now we have to take the limit as it approaches -5 from the left and right, so we have \[\huge \lim_{x \rightarrow -5^{-}} \frac{ 5-(-x) }{ 5+x } = ...\] and \[\huge \lim_{x \rightarrow -5^+} \frac{ 5-x }{ 5+x } = ...\]does that make sense?
Babynini
  • Babynini
Okay, but direct substitution still doesn't work there o.o
Babynini
  • Babynini
They equal, 1 and - 1 respectively?
Zarkon
  • Zarkon
coming from either direction of -5 we have \[\lim_{x \rightarrow -5} \frac{ 5-|x| }{ 5+x } =\lim_{x \rightarrow -5} \frac{ 5-(-x) }{ 5+x } \]
Astrophysics
  • Astrophysics
Yes, thanks @Zarkon
Babynini
  • Babynini
So.. = 1
Astrophysics
  • Astrophysics
Yes
Babynini
  • Babynini
Yay!! ok, perfect. Thanks both of you :)
Astrophysics
  • Astrophysics
Had to note x<0 for both :p
Astrophysics
  • Astrophysics
Slight mistake
Babynini
  • Babynini
Yeah, sorry o.0

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