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anonymous
 one year ago
Can anyone tell me where to find a complete note set on using precise limit definition?
I understand enough to prove limits using it but not enough to answer questions like this.
Let a,L (belongs to) R , and Let f,g,h real value functions. There exist delta such that
f(x)<=g(x)<=h(x) for all x belongs to(adelta, a+delta)
if limit f(x) =L = limit h(x)
x>a x>a
prove that limit g(x)=L
x>a
anonymous
 one year ago
Can anyone tell me where to find a complete note set on using precise limit definition? I understand enough to prove limits using it but not enough to answer questions like this. Let a,L (belongs to) R , and Let f,g,h real value functions. There exist delta such that f(x)<=g(x)<=h(x) for all x belongs to(adelta, a+delta) if limit f(x) =L = limit h(x) x>a x>a prove that limit g(x)=L x>a

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I would get any advanced calculus book. This stuff is very old so they all say pretty much the same thing. I think a good book is the following: An introduction to analysis by Wade.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1It is hard stuff and it takes a while to get comfortable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks. Couldn't find a copy on internet though :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Suppose \(\exists \delta_0 \) s.t. \(xa< \delta \implies f(x)\le g(x) \le h(x)\). Now \(\exists \delta_1>0\) s.t. \(x\in(a\delta_1, x+\delta_1) \implies f(x)L<L\) since \(\lim_{x\rightarrow a}f(x) =\epsilon\). Now we have \(f(x)L<\epsilon \implies \epsilon<f(x)L \implies f(x)>L\epsilon \). Similarly \(\exists \delta_2>0\) s.t. \(x\in(a\delta_2, x+\delta_2) \implies h(x)L<L\) since \(\lim_{x\rightarrow a}h(x) =L\) and \(h(x)L<\epsilon \implies h(x)L<\epsilon \implies h(x)< \epsilon+L\). Now for the proof. Let \(\epsilon > 0 \) be given, define \(\delta =\min\{\delta_0,\delta_1, \delta_2\}\) and suppose \(xa<\delta\) then \(L\epsilon< f(x) \le g(x) \le h(x) < \epsilon+L\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1The result follows because we have \(g(x)L<\epsilon\) when \(x\in (a\delta, a+\delta)\).
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