Delta Epsilon Proof: Prove f(z)/g(z) \rightarrow w1/w2 as z\rightarrow z0

[tylerc1991]

Homework Statement

Define: f(z) $$\rightarrow$$ w1 as z $$\rightarrow$$ z0
and
g(z) $$\rightarrow$$ w2 as z $$\rightarrow$$ z0

prove that f(z)/g(z) $$\rightarrow$$ w1/w2 as z$$\rightarrow$$ z0

The Attempt at a Solution

let $$\epsilon$$ > 0
choose $$\delta$$ > 0 such that:
|f(z) - w1| < ______ (defined later)
|g(z) - w2| < ______ (defined later)

|f(z)/g(z) - w1/w2| = |f(z)/g(z) - f(z)/w2 + f(z)/w2 - w1/w2|
< |f(z)|*1/|g(z)-w2| + 1/|w2|*|f(z) -w1|

this is where I am stuck, I know that you have to make that add up to epsilon but I'm unsure how to pick them so it works out correctly. Any help would be greatly appreciated.

 

[LeonhardEuler]

|f(z)/g(z) - w1/w2| = |f(z)/g(z) - f(z)/w2 + f(z)/w2 - w1/w2|
< |f(z)|*1/|g(z)-w2| + 1/|w2|*|f(z) -w1|

I'm not sure what you did to get to that second line from the first one, but it is surely not what you want because that 1/|g(z)-w2| becomes very large as z$\rightarrow$z0 since the denominator goes to 0. How about getting a common denominator first? Then this will look similar to the problem of proving that the product of the limits is the limit of the product.

 

[tylerc1991]

yes I realized that about 60 seconds after I posted it, but I think I have got it now. I was able to make both sides less than epsilon/2 and I think it works now. thanks for taking a look!