Test Review 5 - limit of a constant sequence

[cmurphy]

I need to prove that the limit of a constant sequence converges, using the definition of a limit.

This is what I have:

Let e > 0 be given.
Then |sn - s| < e
But sn = s for all sn, thus
|s - s| < e
|0| < e
0 < e
Thus N can be any number?

This proof is simple, but I am making it complicated! Please help!

 

[Oxymoron]

Your solution is close but not correct. Remember, the definition of a limit involves [itex]\epsilon[/itex] AND [itex]\delta[/itex]. There is no mention of a [itex]\delta[/itex] in your proof.

 

[HallsofIvy]

Your solution is close but not correct. Remember, the definition of a limit involves [itex]\epsilon[/itex] AND [itex]\delta[/itex]. There is no mention of a [itex]\delta[/itex] in your proof.

No, Oxymoron, there is NO [itex]\delta[/itex] in the proof of a limit of sequenc! That's only for limit of a function in which the variable is takes on continuous values.

CMurphy, your proof is completely correct: N can be taken to be anything. It really is that easy!

 

[Oxymoron]

Sorry guys, didn't realize it was a sequence. Halls is 100% correct, and so are you colleen. However, if it was a function then you need the delta. I hope you realize that you didnt need the delta because you only have a sequence.

 

[Muzza]

However, if it was a function then you need the delta.

That's a truth "with modification", since the epsilon-delta definition of limits (or continuity) can be formulated using only sequences.