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Delta Epsilon Proof

  • Thread starter tylerc1991
  • Start date
  • #1
166
0

Homework Statement



Define: f(z) [tex]\rightarrow[/tex] w1 as z [tex]\rightarrow[/tex] z0
and
g(z) [tex]\rightarrow[/tex] w2 as z [tex]\rightarrow[/tex] z0

prove that f(z)/g(z) [tex]\rightarrow[/tex] w1/w2 as z[tex]\rightarrow[/tex] z0

The Attempt at a Solution



let [tex]\epsilon[/tex] > 0
choose [tex]\delta[/tex] > 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.
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
1
|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[itex]\rightarrow[/itex]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.
 
  • #3
166
0
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!
 

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