## 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(a-delta, a+delta) if limit f(x) =L = limit h(x) x->a x->a prove that limit g(x)=L x->a

1. zzr0ck3r

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.

2. zzr0ck3r

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

3. anonymous

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

4. zzr0ck3r

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 $$L-\epsilon< f(x) \le g(x) \le h(x) < \epsilon+L$$

5. zzr0ck3r

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

6. anonymous

Thnx.:) ok