{key terms/ideas are in
bold, names in
italics.}
1)
Xayma said
brett86 is wrong because he requires
brett86 to approach the point x= 0 from both sides in the use of
l'Hopitals Rule, in which case this shows the
limit does not exist according to
Xayma.
2) i tried to say that all
Xayma proved with that is that the point x=0 is a
discontinuity and that the graph does NOT even exist on the
negative side - this is why an approach from both sides does not work, not because the limit doesn't exist, but because you can't take the limit from both sides since common sense says you can't do that with one whole side missing!
3)
acmilan then introduces the concept of the
epsilon-delta definition, which is a formal test for the existence of a limit at a particular point. however, the epsilon-delta definition takes approaches from both side of the point x= 0, just like
Xayma's method, and consequently shows that a limit at x = 0 does not exist.
4) i tried to argue that, like
Xayma's method, the epsilon-delta definition is inapplicable because of the simple reason that you cannot take an approach from both side of x=0 since one side is just simply missing.
5) i found a formal
mathworld site which stipulates the use of the epsilon-delta definition as a test for
continuity about a point of a graph/function. the limit test, of course, itself being one of the three requirements that defines a locally continuous region on a curve. hence, the implication being that you cannot use the epsilon-delta definition to test for a limit about a discontinuous point - which x = 0 on y=x^x is. {especially with one whole side missing} x= 0 is also a
critical point.
6) the website suggested the use of l'Hopitals Rule under the circumstances where the epsilon-delta definition does not work (ie. in this case), which brings us back to the initial use of l'Hopitals Rule by
brett86 that shows 0^0 has a limit of 1. however, knowing that the use of l'Hopitals Rule is required now only because we can't use the epsilon-delta definition means that an approach from both sides is no longer necessary (otherwise it'd be no different than the epsilon-delta). and so approaching from the most appropriate side, ie. where the graph actually exists, brings us back to the point
brett86 already made: 0^0 has a limit of 1.
7)
Xayma introduces the concept of a
complex limit. however, the graph is on a
real number plane, NOT an
Argand plane, and so any complex limits is irrelevant since the limit of the curve/graph is real. this introduction of a complex limit only "sidesteps" the real issue of the existence of a limit at x=0 for
Xayma, since this is not a real argument/solution to the issue.
8)
Slide_Rule suggests the inapproriateness of using l'Hopitals Rule on the function y= x^x as x --> 0 .
9)
brett86 defends his use of l'Hopitals Rule successfully with support from mathematical software, such as
Maple - which all agree the limit of 0^0 to be = 1. In addition, many posts ago, i previously posted a graph of the function y=x^x near x=0, as drawn by
Mathematica, which supports
brett86's conclusion that 0^0 has a limit of 1.
10)
acmilan questions the continuity of y=(x^2 -4)/(x +2) at the point where x= 2, saying that it is indeed continuous at x=2 and that the epsilon-delta definition can thus be applied.
acmilan further quotes the limit at x=2 as L =2.
11) i am pretty sure, as is
brett86, that at x=2, the graph of y=(x^2 -4)/(x +2) is NOT continuous... meaning that the epsilon-delta definition cannot be applied, further implying, according to
Slide_Rule, that a limit at x=2 does not actually exist.
12) most parties agree now that the limit at x=2 on y=(x^2 -4)/(x +2) does in fact exist, and it is L = 2 as
acmilan had quoted originally.
13)
Xayma re-introduces the complex limit.
14)
brett86 questions, like i did, the validity of introducing the complex limit again. the reason being we are looking at a graph on the real number plane - nothing to do with the Argand plane (they are separate planes!).
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