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ParthKohli

  • 2 years ago

Is there a removable discontinuity when a limit exists?

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  1. abb0t
    • 2 years ago
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    a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point. |dw:1356113778802:dw|

  2. ParthKohli
    • 2 years ago
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    What if there's a hole at a point and the limit *is* the hole? What discontinuity is that called?

  3. TuringTest
    • 2 years ago
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    ...but whether or not the unction value exists, for a removeable discontinuity we have that\[\lim_{x\to a^1}f(x)=\lim_{x\to a^+}f(x)\neq f(a),\pm\infty\]i.e. the limit must exist for a removeable discontinuity.

  4. abb0t
    • 2 years ago
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    Step Continuity?

  5. ParthKohli
    • 2 years ago
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    Oh, I get it.

  6. TuringTest
    • 2 years ago
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    the thing you describe parth is a removable discontinuity

  7. ParthKohli
    • 2 years ago
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    Ah, okay :)

  8. TuringTest
    • 2 years ago
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    the limit exists (is finite) but is not the same as the value of the function at that point. the actual value of the function at that point can be anything... even f(a)=infinity, it would still be removable

  9. ParthKohli
    • 2 years ago
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    Got it, so the discontinuity in a function\[f(x) = {{\rm foo} \over x + 2}\]is a removable discontinuity right?

  10. TuringTest
    • 2 years ago
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    no, because \[\lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x)\]

  11. TuringTest
    • 2 years ago
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    I mean \(x\to-2\)

  12. TuringTest
    • 2 years ago
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    also, both limits do not exist; for a removable discontinuity the limit at the discontinuity must exist ( be finite and equal on both sides)

  13. TuringTest
    • 2 years ago
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    here \(x\to2^-\implies f(x)\to\infty,~~~x\to2^+\implies f(x)\to-\infty\)

  14. ParthKohli
    • 2 years ago
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    Ah! Okay, so the limit doesn't exist.

  15. TuringTest
    • 2 years ago
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    so the limits are neither equal on both sides, or even the same sign!

  16. TuringTest
    • 2 years ago
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    correct, for many reasons it doesn't exist

  17. TuringTest
    • 2 years ago
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    it is an essential discontinuity

  18. ParthKohli
    • 2 years ago
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    hmm

  19. TuringTest
    • 2 years ago
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    |dw:1356114605529:dw|

  20. ParthKohli
    • 2 years ago
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    I see :)

  21. TuringTest
    • 2 years ago
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    as \(x\to-2^+\) we are going this way off to infinity|dw:1356114734440:dw|

  22. ParthKohli
    • 2 years ago
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    Yup, I see it now :) thanks

  23. TuringTest
    • 2 years ago
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    that alone is enough to say that the discontinuity is essential, and that the limit does not exist at x=-2 but still we have the other argument as well. from the left we have the opposite case|dw:1356114780902:dw|

  24. ParthKohli
    • 2 years ago
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    Aww, no one cares about the vertical asymptote \(x = -2\)... give it some love :)

  25. TuringTest
    • 2 years ago
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    forever alone I guess :P

  26. ParthKohli
    • 2 years ago
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    lol

  27. ParthKohli
    • 2 years ago
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    Thanks @TuringTest :)

  28. TuringTest
    • 2 years ago
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    welcome :D

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