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Babynini

  • one year ago

Limits of absolute values

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  1. Babynini
    • one year ago
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  2. Jhannybean
    • one year ago
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    @Astrophysics

  3. Astrophysics
    • one year ago
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    Ok lets first note, |dw:1443917561077:dw| if we plug the limit we will get 0/0 correct?

  4. Babynini
    • one year ago
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    Right

  5. Astrophysics
    • one year ago
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    All we really need to know is |dw:1443917619092:dw| so we have to break it up as such

  6. Astrophysics
    • one year ago
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    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?

  7. Babynini
    • one year ago
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    Okay, but direct substitution still doesn't work there o.o

  8. Babynini
    • one year ago
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    They equal, 1 and - 1 respectively?

  9. Zarkon
    • one year ago
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    coming from either direction of -5 we have \[\lim_{x \rightarrow -5} \frac{ 5-|x| }{ 5+x } =\lim_{x \rightarrow -5} \frac{ 5-(-x) }{ 5+x } \]

  10. Astrophysics
    • one year ago
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    Yes, thanks @Zarkon

  11. Babynini
    • one year ago
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    So.. = 1

  12. Astrophysics
    • one year ago
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    Yes

  13. Babynini
    • one year ago
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    Yay!! ok, perfect. Thanks both of you :)

  14. Astrophysics
    • one year ago
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    Had to note x<0 for both :p

  15. Astrophysics
    • one year ago
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    Slight mistake

  16. Babynini
    • one year ago
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    Yeah, sorry o.0

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