anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

zzr0ck3r
  • 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.
zzr0ck3r
  • zzr0ck3r
It is hard stuff and it takes a while to get comfortable
anonymous
  • anonymous
Thanks. Couldn't find a copy on internet though :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zzr0ck3r
  • 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. \(x\in(a-\delta_1, x+\delta_1) \implies |f(x)-L|L-\epsilon \). Similarly \(\exists \delta_2>0\) s.t. \(x\in(a-\delta_2, x+\delta_2) \implies |h(x)-L| 0 \) be given, define \(\delta =\min\{\delta_0,\delta_1, \delta_2\}\) and suppose \(|x-a|<\delta\) then \(L-\epsilon< f(x) \le g(x) \le h(x) < \epsilon+L\)
zzr0ck3r
  • zzr0ck3r
The result follows because we have \(|g(x)-L|<\epsilon\) when \(x\in (a-\delta, a+\delta)\).
anonymous
  • anonymous
Thnx.:) ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.