Delta Epsilon Limit Proof - f(x,y) = xy

  • Thread starter JeffNYC
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  • #1
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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
 

Answers and Replies

  • #2
HallsofIvy
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Yes, you want [itex]|xy- ab|< \epsilon[/itex] . Now, xy- ab= xy- ay+ ay- ab= y(x-a)+ a(y-b) so [itex]|xy- ab|\le |y||x-a|+ |a||y- b|[/itex]. Since y converges to b, you can make |y-b| as small as you like so you can make it less than [itex]\epsilon/2a[/itex]. The other term is a bit tricky. You need to make |y||x-a| less than [itex]\epsilon/2[/itex] but y is a variable. However, since y converges to b, you can make |y| less than |b|+1 so you can use that instead: since x converges to a, you can make [itex]|x-a|\le \epsilon/(|b|+ 1)< \epsilon/|y|[/itex].
 
  • #3
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Any idea what's happening with the latex output?
 
  • #4
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( I assume a,b ≠ 0 )

| x - a | < ε' hence | x∙y - a∙y | < ε' ∙ | y |
| y - b | < ε' hence | y∙a - b∙a | < ε' ∙ | a |

you sum both inequalities

| x∙y - a∙y | + | y∙a - b∙a | < ε' ∙ | y | + ε' ∙ | a |

you use the triangle inequality

| x∙y - a∙y | + | y∙a - b∙a | > | x∙y - a∙y + y∙a - b∙a | = | x∙y - a∙b |

define ε = ε' ∙ | y | + ε' ∙ | b |

now putting everything together yields : ...
 
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  • #5
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maybe multiplying works too?

-d1 + a < x < d1 + a
-d2 + b < y < d2 + b

=>
since -d1d2 < d1d2,

-d1d2-d1b-d2a < d1d2-d1b-d2a < xy - ab < d1d2 + d1b + d2a

then,

abs(xy - ab) < d1d2 + d1b + d2a

since a and b are fixed, as d1 and d2 go to zero, the abs goes to zero.

[edit: hm. no, the sign on the multiplication makes this an error.]
 
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