Question

How do you prove that the limit of `x+2y= 1` as (x,y) approaches (1,0) using the epsilon delta proof?

Answer 1

Set `y=f(x)=1/2*(1-x)` hence we want show that

`lim_(x->1) f(x)=0`

Suppose we are given `ε > 0`. We then wish to find `δ > 0` such
that for any `x ∈ R`, `0 < |x − 1| < δ` implies `|1/2(1-x) − 0| < ε` or
`1/2*|(x-1)| < ε` or `|(x-1)| <2* ε` .

If we choose `δ=min{2*ε,1}` would be an alright choice for `δ`

Answer 2

If I ever did this it was years ago. I did look at some resources for `epsilon`-`delta` definitions of such limits and here is the result.

Explanation:

Given `epsilon > 0`, choose `delta = epsilon/3`. (Note that `delta` is positive.)

Now if `(x,y)` is within `delta` of `(1,0)` but distinct from `(1,0)`,

then `0 < sqrt((x-1)^2+y^2) < delta`.

So `sqrt((x-1)^2) = abs(x-1) < delta` and `sqrt(y^2) = abs(y) < delta`.

Consequently,

`abs((x+2y)-1)) = abs((x-1)+2y)`

`<= abs(x-1)+2abs(y)`

`< delta + 2delta`

`= epsilon/3 + 2epsilon/3`

`= epsilon`.