Test Review 5 - limit of a constant sequence

cmurphy
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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!
 
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Your solution is close but not correct. Remember, the definition of a limit involves \epsilon AND \delta. There is no mention of a \delta in your proof.
 
Oxymoron said:
Your solution is close but not correct. Remember, the definition of a limit involves \epsilon AND \delta. There is no mention of a \delta in your proof.

No, Oxymoron, there is NO \delta 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!
 
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.
 
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.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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