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## Main Question or Discussion Point

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