- #1
JeffNYC
- 26
- 0
How does one prove:
limit xy = ab
x-> a
y -> b
using the precise definition of a limit?
My attempt:
|xy-ab|<ϵ
for:
0<|x-a|<δ/2
0<|y-b|<δ/2
it follows that:
δ/2-a <x< δ/2+a
δ/2-b <y< δ/2+b
then:
(δ/2-a)(δ/2-n) < xy < (δ/2+a)(δ/2+b)
(δ^2/4-aδ/2-bδ/2 +ab) <xy< (δ^2/4+aδ/2+bδ/2+ab )
So -
|xy-ab|< (δ^2/4+aδ/2+bδ/2)
and I need to make
|xy-ab|<ϵ
At this point I feel a bit lost - anyone have any ideas for this proof?
Thanks,
Jeff
limit xy = ab
x-> a
y -> b
using the precise definition of a limit?
My attempt:
|xy-ab|<ϵ
for:
0<|x-a|<δ/2
0<|y-b|<δ/2
it follows that:
δ/2-a <x< δ/2+a
δ/2-b <y< δ/2+b
then:
(δ/2-a)(δ/2-n) < xy < (δ/2+a)(δ/2+b)
(δ^2/4-aδ/2-bδ/2 +ab) <xy< (δ^2/4+aδ/2+bδ/2+ab )
So -
|xy-ab|< (δ^2/4+aδ/2+bδ/2)
and I need to make
|xy-ab|<ϵ
At this point I feel a bit lost - anyone have any ideas for this proof?
Thanks,
Jeff