Question

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(a-delta, a+delta)

if limit f(x) =L = limit h(x)
x->a x->a
prove that limit g(x)=L
x->a

Answer 1

I 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.

Answer 2

It is hard stuff and it takes a while to get comfortable

Answer 3

Thanks. Couldn't find a copy on internet though :)

Answer 4

Suppose \(\exists \delta_0 \) s.t. \(|x-a|< \delta \implies f(x)\le g(x) \le h(x)\). 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 \(|x-a|<\delta\) then \(L-\epsilon< f(x) \le g(x) \le h(x) < \epsilon+L\)

Answer 5

The result follows because we have \(|g(x)-L|<\epsilon\) when \(x\in (a-\delta, a+\delta)\).

Answer 6

Thnx.:) ok