Multivariable limits with epsilon delta help?

2016-02-07 8:24 pm
更新1:

Sorry, this was what I wanted to ask: I need to prove that the limit of f(x)= xy is 8 as (x,y) approaches (-1,8) using the definition of limit. How do you do this?

更新2:

The limit is actually -8 not 8.

回答 (1)

2016-02-08 4:23 am
✔ 最佳答案
Given ε > 0, we need to find ε > 0 such that
0 < √((x+1)² + (y-8)²) < δ ==> |xy - (-8)| < ε.

To this end, note that
|xy - (-8)|
= |((x+1) - 1)((y-8) + 8) + 8|, via 'adding 0' cunningly
= |((x+1)(y-8) + 8(x+1) - (y-8) - 8) + 8|, via FOIL
= |(x+1)(y-8) + 8(x+1) - (y-8)|
≤ |x+1| |y-8| + 8 |x+1| + |y-8|, via triangle inequality
= √(x+1)² √(y-8)² + 8√(x+1)² + √(y-8)²
≤ √((x+1)² + (y-8)²) √((x+1)² + (y-8)²) + 8√((x+1)² + (y-8)²) + √((x+1)² + (y-8)²)
< √((x+1)² + (y-8)²) + 8√((x+1)² + (y-8)²) + √((x+1)² + (y-8)²),
assuming that √((x+1)² + (y-8)²) < 1
= 10√((x+1)² + (y-8)²).

So given ε > 0, let δ = min{1, ε/10}. Then, 0 < √((x+1)² + (y-8)²) < δ
==> |xy - (-8)| < 10√((x+1)² + (y-8)²) < 10(ε/10) = ε.

I hope this helps!


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