- #1
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could anyone produce a proof showing why a sequence can't have more than 1 unique limit?
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In summary, the conversation discusses the proof that a sequence cannot have more than one unique limit. The proof involves using the definition of a limit as well as a proof by contradiction. The key idea is that if a sequence has two limits, then it must converge to the same point, leading to a contradiction. The conversation also mentions the preference for a geometric understanding of convergence over epsilon-delta arguments.
- #2
quasar987
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Suppose {x_n} has two limits, say x and y. Then this means that
1) for all e > 0, there exists and N_1 > 0 s.t. n > N_1 ==> |x_n - x| < e
2) for all e > 0, there exists and N_2 > 0 s.t. n > N_2 ==> |x_n - y| < e
It follows that for n > max{N_1, N_2}, |x_n - x| < e and |x_n - y| < e.
Therefor, for n > max{N_1, N_2}, |x - x_n + x_n - y| [itex]\leq[/itex] |x - x_n| + |x_n - y| = |x - x_n| + |x_n - y| < e + e = 2e.
But |x - x_n + x_n - y| = |x - y|. So this last inequality writes, in limit notation,
[tex]\lim_{n \rightarrow \infty} (x - y) = 0[/tex]
but x - y = a constant. We know that the limit of a constant is the constant itself. So [itex]\lim_{n \rightarrow \infty} (x - y) = x - y [/itex]. So x - y = 0.
Edit: So x = y. [itex]\blacksquare[/itex] (P.S. No, I don't think you're that dumb, I just like my proofs complete to the last drop... kind of a obsesso-maniacal thing I have
)
1) for all e > 0, there exists and N_1 > 0 s.t. n > N_1 ==> |x_n - x| < e
2) for all e > 0, there exists and N_2 > 0 s.t. n > N_2 ==> |x_n - y| < e
It follows that for n > max{N_1, N_2}, |x_n - x| < e and |x_n - y| < e.
Therefor, for n > max{N_1, N_2}, |x - x_n + x_n - y| [itex]\leq[/itex] |x - x_n| + |x_n - y| = |x - x_n| + |x_n - y| < e + e = 2e.
But |x - x_n + x_n - y| = |x - y|. So this last inequality writes, in limit notation,
[tex]\lim_{n \rightarrow \infty} (x - y) = 0[/tex]
but x - y = a constant. We know that the limit of a constant is the constant itself. So [itex]\lim_{n \rightarrow \infty} (x - y) = x - y [/itex]. So x - y = 0.
Edit: So x = y. [itex]\blacksquare[/itex] (P.S. No, I don't think you're that dumb, I just like my proofs complete to the last drop... kind of a obsesso-maniacal thing I have
)
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- #3
matt grime
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Another one, by contradiction (which is essentially the same as the previous one) is: if x_n tends to x and x_n tends to y. If x=/=y, then let e=|x-y|/3, then for all n sufficently large we have
|x_n - x| < e , ie all the x_n are in the interval (x-e,x+e) for n >N
and
|x_n - y| <e, ie all the x_n are in the interval (y-e,y+e) for n>N
but the two intervals (x-e,x+e) and (y-e,y+e) are disjoint so it is impossible for all the x_n for n>N to be in two disjoint intervals. a contradiction. the only assumption was that x=/=y, so it must be that x=y.
Personally I prefer this one (even though it is an unnecessary proof by contradiction) because it makes you think visually (or geometrically) about what it means for a seqqunce to converge: for all open intervals about x (or open "balls" in higher dimensions) that if we throw away some finite number of terms at the start of the sequence then all of the ones that are left are inside the inteval or ball. it gets awayy from the epsilon delta arguments, and that has to be a good thing.
|x_n - x| < e , ie all the x_n are in the interval (x-e,x+e) for n >N
and
|x_n - y| <e, ie all the x_n are in the interval (y-e,y+e) for n>N
but the two intervals (x-e,x+e) and (y-e,y+e) are disjoint so it is impossible for all the x_n for n>N to be in two disjoint intervals. a contradiction. the only assumption was that x=/=y, so it must be that x=y.
Personally I prefer this one (even though it is an unnecessary proof by contradiction) because it makes you think visually (or geometrically) about what it means for a seqqunce to converge: for all open intervals about x (or open "balls" in higher dimensions) that if we throw away some finite number of terms at the start of the sequence then all of the ones that are left are inside the inteval or ball. it gets awayy from the epsilon delta arguments, and that has to be a good thing.
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- #4
- 668
- 68
Thanks guys, exactly the notation and process I was looking for
FAQ: Sequence can't have more than 1 unique limit?
What is a unique limit in a sequence?
A unique limit in a sequence refers to the value that the terms in the sequence approach or tend to as the sequence progresses. In other words, it is the value that the terms in the sequence get closer and closer to, but never actually reach.
Can a sequence have more than one unique limit?
No, a sequence cannot have more than one unique limit. This is because a sequence can only approach or tend to one specific value as it progresses. If there are multiple unique limits, then the sequence would not have a clear and definite value that it is approaching.
What happens if a sequence has more than one unique limit?
If a sequence has more than one unique limit, then it is not considered a valid sequence. This is because it violates the definition of a sequence, which states that a sequence must have a single value that it approaches as it progresses.
Are there any exceptions to the rule that a sequence can't have more than one unique limit?
No, there are no exceptions to this rule. This rule is a fundamental property of sequences and cannot be violated. If a sequence has more than one unique limit, then it is not considered a valid sequence.
Why is it important for a sequence to have only one unique limit?
Having only one unique limit in a sequence allows us to make precise and accurate predictions about the behavior of the sequence. It also helps us to understand the underlying patterns and relationships within the sequence. If a sequence has more than one unique limit, then it becomes much more difficult to analyze and make predictions about.