Is there a removable discontinuity when a limit exists?

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Is there a removable discontinuity when a limit exists?

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

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