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Trouble understanding

by sutupidmath
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sutupidmath
#1
Feb23-08, 09:03 PM
P: 1,635
Well there is this theorem and i am having a little trouble understanding a part of it:

-If the function u=g(x) has Uo as the limit at the limit point a (x-->a), but there exists some delta1 neighbourhood of a, such that for 0<Ix-aI<delta1 we have u=g(x)=/=(not equal) Uo, and if the function y=f(u) has b as the limit at the limit point Uo(u-->Uo), then the compound function y=f[g(x)] has b as the limit at the limit point a. That is:

If lim(x-->a) g(x)=Uo ( u=g(x)=/=Uo for 0<Ix-aI<delta1 )

and lim(u-->Uo)f(u)=b then

lim(x-->a)f[g(x)]=b.

The part that i do not fully understand, since i am not able to apply it on problems is the additional requirement on the first part of this theorem If the function u=g(x) has Uo as the limit at the limit point a (x-->a), but there exists some delta1 neighbourhood of a, such that for 0<Ix-aI<delta1 we have u=g(x)=/=(not equal) Uo

My book does not provide any examples at all to ilustrate what happenes if this requirement is not fullfilled, and i am having trouble coming up with any examples that would illustrate this point. SO if someone could post some examples and point out why this requirement is crucial for the theorem to hold, i would really appreciate it.

SO if you can explain this a little, and throw some examples illustrating this i would really appreciate it.

thnx in advance
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mathman
#2
Feb24-08, 04:36 PM
Sci Advisor
P: 6,040
I can't fully grasp the point that is being made in the text. However, you could examine the implications of g(x) being a constant. It could be an example where f(Uo) is not b, i.e. there is some sort of singular point at Uo.
sutupidmath
#3
Feb24-08, 04:44 PM
P: 1,635
yeah i know that if this :but there exists some delta1 neighbourhood of a, such that for 0<Ix-aI<delta1 we have u=g(x)=/=(not equal) Uo

would not hold, say that if there exists a delta1 neighbourhood of a, such that for every x from that neighbourhood we have u=g(x)=Uo, then we would not be sure at all that this would exist

lim(u-->Uo)f(u)=b then
since from the def, we would have

for every epsylon there exists some delta such that whenever 0<Iu-UoI<delta, we have
If(u)-bI<epsylon,
so if we had this u=g(x)=Uo for some x in a neighbourhood of a, then we would not be sure that the lim(u-->Uo)f(u)=b exists since it would not fullfile the requirement 0<Iu-UoI<delta, so it is possible that the limit at Uo not to exist.
But i just need some examples to illustrate this.?

Mystic998
#4
Feb24-08, 07:49 PM
P: 206
Trouble understanding

I think the point of saying that such a neighborhood shouldn't exist is that it's possible for f(u) to be something like 1/(u - Uo). But if u(x) is identically Uo in a neighborhood of a, the limit can't exist. But I think if we assume that f(u) has at worst a removable singularity at Uo, then you don't need the assumption that such a neighborhood about a doesn't exist. But don't quote me on that.


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