Question

Prove the theorem that if limx→d h(x) = P and limx→d k(x) = Q and h(x) ≥ k(x) for all x in an open interval containing d, then P ≥ Q by using the formal definition of the limit, showing all work.

Answer 1

limx→d h(x) = P mean h(d)=P
limx→d k(x) = Q mean k(d)=Q
but h(x) ≥ k(x) for all x in an open interval containing d
thus h(d) ≥ k(d)
that implies
P≥Q

Answer 2

Thanks but I have to prove using the epsilon delta definition of the limit. Though your proof is very nice and to the point.

Answer 3

ok il prove it to u using epsilon delta definition of the limit..
brb

Answer 4

limx→d (h(x)-k(x)) = P -Q
so u need to prove P-Q ≥0.
using the def
for each epsilon >0 there exist delta >0 ,S.t
0<|x-d|<delta
|(h(x)-k(x))-(P-Q)|<epsilon
but h(x)-k(x)≥0 (h(x)-k(x))-(P-Q)≥-(P-Q)
mmm... ok il try to continue that ltr :\

Answer 5

that is great start but I keep getting stuck also