- #1
lionel_hutz
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Homework Statement
Using the delta-epsilon definition of limits, prove that of lim f(x) = l and lim f(x) =m, then l=m
Homework Equations
Delta-epsilon definiition of the limit of f(x), as x approaches a:
For all e>0, there is a d s.t if for all x, |x-a|<d, then |f(x) -l|<e
The Attempt at a Solution
Suppose not. Then
(1): For all e>0, there is a d1 s.t. |f(x) - l|<e, provided |x-a|<d1 for all x
(2): For all e>0, there is a d2 s.t. |f(x) - m|<e, provided |x-a|<d2 for all x
I'm getting stuck at finding the delta that works for both of them. I see that the next step is to pick d=min(d1, d2), but I'm confused as to why this is the case
For instance, if d1=1, and d2=2, I don't see how d=1 would satisfy (2) above