Question:
Prove there is a sequence {xn}, satisfying lim(xn) =0 as n goes to infinity, such that lim f(xn) is not L.?
Philip
2012-10-10 11:26:14 UTC
Suppose a function f(x) is defined on (-1,1) and lim f(x) is not equal to L as x goes to 0.
Prove that there exists a sequence of numbers {xn} satisfying lim xn as n goes to infinity = 0
such that lim f(xn) is not equal to L as n goes to infinity.
Three answers:
?
2012-10-10 11:33:25 UTC
Define f(x) = 0 if and only if x = 0, but equals to 1/2 if x is any other value on (-1, 1)



Now, consider the sequence of xn = abs (x^n) for x in (-1, 1). Observe that xn approaches 0 as n >>0, but f(xn) = 1/2, so it's limit = 1/2
No Mythology
2012-10-12 04:59:40 UTC
Since the limit as x->0 f(x) is different from L, there exists ε_0 >0 such that for every δ > 0 there exists x such that



0 < |x - 0| < δ but |f(x) - L| ≥ ε_0.



Now consider the sequence of delta values δ_n = 1/n. By the above, we can construct a sequence {x_n} in the following way:



For each n, choose x_n such that



0 < |x_n - 0| < δ_n and |f(x_n) - L| ≥ ε_0.



That is, lim(n->∞) f(x_n) ≠ L. It only remains to show that x_n -> 0.



Notice that for any ε > 0, we can choose N such that N ≥ 1/ε. Then for every n ≥ N we have



|x_n - 0| < 1/n ≤ ε.



Hence x_n -> 0. So we've constructed a sequence {x_n} convergent to 0 and such that f(x_n) does not converge to L.
bovee
2016-07-30 21:00:34 UTC
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