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ParthKohli
Group Title
Is there a removable discontinuity when a limit exists?
 one year ago
 one year ago
ParthKohli Group Title
Is there a removable discontinuity when a limit exists?
 one year ago
 one year ago

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abb0t Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
What if there's a hole at a point and the limit *is* the hole? What discontinuity is that called?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
...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.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
Step Continuity?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Oh, I get it.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the thing you describe parth is a removable discontinuity
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Ah, okay :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Got it, so the discontinuity in a function\[f(x) = {{\rm foo} \over x + 2}\]is a removable discontinuity right?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no, because \[\lim_{x\to0^}f(x)\neq\lim_{x\to0^+}f(x)\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I mean \(x\to2\)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
also, both limits do not exist; for a removable discontinuity the limit at the discontinuity must exist ( be finite and equal on both sides)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
here \(x\to2^\implies f(x)\to\infty,~~~x\to2^+\implies f(x)\to\infty\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Ah! Okay, so the limit doesn't exist.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so the limits are neither equal on both sides, or even the same sign!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
correct, for many reasons it doesn't exist
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it is an essential discontinuity
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1356114605529:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
I see :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
as \(x\to2^+\) we are going this way off to infinitydw:1356114734440:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Yup, I see it now :) thanks
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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 casedw:1356114780902:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Aww, no one cares about the vertical asymptote \(x = 2\)... give it some love :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
forever alone I guess :P
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Thanks @TuringTest :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
welcome :D
 one year ago
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