Sequences & Limits: Finding the Limit as n->inf

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SUMMARY

The discussion focuses on finding the limit of the sequence defined by x_{n}(t) as n approaches infinity. The sequence is piecewise defined: for 0 ≤ t ≤ 1/n, x_{n}(t) = nt, and for 1/n ≤ t ≤ 1, x_{n}(t) = 1/(nt). As n increases, for any fixed t0 in the interval [0, 1], the limit converges to 0. Participants clarify that regardless of the choice of t0, the function approaches 0 as n approaches infinity.

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Homework Statement



x_{n}(t) \left\{\begin{array}{cc}nt,&amp;\mbox{ if }<br /> 0\leq t \leq \frac{1}{n}\\ \frac{1}{nt} &amp; \mbox{ if } \frac{1}{n}\leq t \leq 1 \end{array}\right.


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The Attempt at a Solution



Can someone help me get started finding the limit as n -> inf? I've never taken the limit of a sequence that has such a dependence on t.

For t in [0, (1/n)], the values of the sequence will range between 0 and 1, and for t in [(1/n),1], the values will range between 0 and 1 as well. It doesn't really matter how large you take n...
 
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Pick a fixed x0 in [0,1] and think about limit x_n(x0) as n->infinity. If x0 is not zero there is always an N>0 such that 1/N<x0. That means for all n>N the definition of x_n(x0) is 1/(n*x0). What's the limit at x0?
 
What do you mean by pick and x0? You mean, pick a t0?
 
Somefantastik said:
What do you mean by pick and x0? You mean, pick a t0?

t0, x0 whatever. Sure, call the point t0 if you want.
 
How about Alfred? Anyway, I think I got what you are saying. No matter what your choice for t, this function will merge to 0 as n -> inf.

thank you for your time.
 

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