ParthKohli
  • ParthKohli
Is there a removable discontinuity when a limit exists?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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abb0t
  • abb0t
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|
ParthKohli
  • ParthKohli
What if there's a hole at a point and the limit *is* the hole? What discontinuity is that called?
TuringTest
  • TuringTest
...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.

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abb0t
  • abb0t
Step Continuity?
ParthKohli
  • ParthKohli
Oh, I get it.
TuringTest
  • TuringTest
the thing you describe parth is a removable discontinuity
ParthKohli
  • ParthKohli
Ah, okay :)
TuringTest
  • TuringTest
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
ParthKohli
  • ParthKohli
Got it, so the discontinuity in a function\[f(x) = {{\rm foo} \over x + 2}\]is a removable discontinuity right?
TuringTest
  • TuringTest
no, because \[\lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x)\]
TuringTest
  • TuringTest
I mean \(x\to-2\)
TuringTest
  • TuringTest
also, both limits do not exist; for a removable discontinuity the limit at the discontinuity must exist ( be finite and equal on both sides)
TuringTest
  • TuringTest
here \(x\to2^-\implies f(x)\to\infty,~~~x\to2^+\implies f(x)\to-\infty\)
ParthKohli
  • ParthKohli
Ah! Okay, so the limit doesn't exist.
TuringTest
  • TuringTest
so the limits are neither equal on both sides, or even the same sign!
TuringTest
  • TuringTest
correct, for many reasons it doesn't exist
TuringTest
  • TuringTest
it is an essential discontinuity
ParthKohli
  • ParthKohli
hmm
TuringTest
  • TuringTest
|dw:1356114605529:dw|
ParthKohli
  • ParthKohli
I see :)
TuringTest
  • TuringTest
as \(x\to-2^+\) we are going this way off to infinity|dw:1356114734440:dw|
ParthKohli
  • ParthKohli
Yup, I see it now :) thanks
TuringTest
  • TuringTest
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|
ParthKohli
  • ParthKohli
Aww, no one cares about the vertical asymptote \(x = -2\)... give it some love :)
TuringTest
  • TuringTest
forever alone I guess :P
ParthKohli
  • ParthKohli
lol
ParthKohli
  • ParthKohli
Thanks @TuringTest :)
TuringTest
  • TuringTest
welcome :D

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