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abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0a 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
 2 years ago
Best ResponseYou've already chosen the best response.2What if there's a hole at a point and the limit *is* the hole? What discontinuity is that called?

TuringTest
 2 years ago
Best 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.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1the thing you describe parth is a removable discontinuity

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1the 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
 2 years ago
Best ResponseYou've already chosen the best response.2Got it, so the discontinuity in a function\[f(x) = {{\rm foo} \over x + 2}\]is a removable discontinuity right?

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1here \(x\to2^\implies f(x)\to\infty,~~~x\to2^+\implies f(x)\to\infty\)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2Ah! Okay, so the limit doesn't exist.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1so the limits are neither equal on both sides, or even the same sign!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1correct, for many reasons it doesn't exist

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1it is an essential discontinuity

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1356114605529:dw

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2Yup, I see it now :) thanks

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1that 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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1forever alone I guess :P

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2Thanks @TuringTest :)
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