No problem, glad I could help!

[lionel_hutz]

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

 

[VantagePoint72]

Given some $\delta$ that works in an epsilon-delta proof, you can always take a smaller delta, $\delta'$, instead (since $|x-a|<\delta' < \delta \rightarrow |x-a| < \delta$). So if $\delta_1$ works for (1) and $\delta_2$ works for (2), then $min(\delta_1,\delta_2)$ works for both.

 

[lionel_hutz]

Given some $\delta$ that works in an epsilon-delta proof, you can always take a smaller delta, $\delta'$, instead (since $|x-a|<\delta' < \delta \rightarrow |x-a| < \delta$). So if $\delta_1$ works for (1) and $\delta_2$ works for (2), then $min(\delta_1,\delta_2)$ works for both.

Alright, I see. I was thinking max and confusing myself, lol. Thanks!