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hawkfalcon

  • 2 years ago

limits:O

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  1. hawkfalcon
    • 2 years ago
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    \[\lim_{x \rightarrow 5} \frac{ x-5 }{ \left| x-5 \right| }\]

  2. vf321
    • 2 years ago
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    Well, do you know that\[\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a^+}f(x)\cap\lim_{x\rightarrow a^-}f(x)\]I.e., for 1 variable, the bilateral limit exists iff both lateral limits exist and are the same?

  3. hawkfalcon
    • 2 years ago
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    Erm, yes.

  4. vf321
    • 2 years ago
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    Okay. Then coming in from the right at 5, what is your limit equal?\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=?\]

  5. hawkfalcon
    • 2 years ago
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    0/0

  6. hawkfalcon
    • 2 years ago
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    oh wait. \[infinity\]

  7. hawkfalcon
    • 2 years ago
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    would it be DNE?

  8. vf321
    • 2 years ago
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    The right-side limit is not DNE. When solving limits you can only say that\[\lim_{x\rightarrow a}f(x)=f(a)\] if f(a) exists.

  9. vf321
    • 2 years ago
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    Think about it. We're approaching the limit from the right. When x = 6, what's the limit? When x = 5.01, what's the limit?

  10. hawkfalcon
    • 2 years ago
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    okay. Wait each side would be infinity right?

  11. vf321
    • 2 years ago
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    Try plugging in values I showed you. Also, a note for the prior statement I made: f has to be continuous.

  12. hawkfalcon
    • 2 years ago
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    at 6 it would be -1/2

  13. hawkfalcon
    • 2 years ago
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    right?:3

  14. hawkfalcon
    • 2 years ago
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    I read that it has to be either DNE, infinity, or negative infinity

  15. hawkfalcon
    • 2 years ago
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    @vf321 sorry, i'm confused :(:#

  16. vf321
    • 2 years ago
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    Lets try this again.\[\frac{6-5}{|6-5|}=1\] how on earth did u get 1/2?

  17. hawkfalcon
    • 2 years ago
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    I was accidentally looking at a different problem:3

  18. vf321
    • 2 years ago
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    okay. now try 5.0001

  19. hawkfalcon
    • 2 years ago
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    But okay, then 5.0001 = 1 too.

  20. vf321
    • 2 years ago
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    okay. try 5.000000000000000001

  21. hawkfalcon
    • 2 years ago
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    Anything >5 will be 1

  22. vf321
    • 2 years ago
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    good. In fact, you can say\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=1\]

  23. vf321
    • 2 years ago
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    Now try\[\lim_{x\rightarrow 5^-}f(x)=?\]From the left, on your own.

  24. hawkfalcon
    • 2 years ago
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    o.o okay that makes sense. Let me try...

  25. hawkfalcon
    • 2 years ago
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    -1

  26. hawkfalcon
    • 2 years ago
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    So it's DNE, because they are different.

  27. vf321
    • 2 years ago
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    yes! and, if you were to graph the function on a CAS (or Wolfram|Alpha), you'd see there's a jump discontinuity at 0.

  28. hawkfalcon
    • 2 years ago
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    Do I write that?:O

  29. vf321
    • 2 years ago
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    no but its for ur understanding.

  30. hawkfalcon
    • 2 years ago
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    Oh okay, thank you:)

  31. vf321
    • 2 years ago
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    remeber to close the question.

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