Cauchy Sequence Convergence in the Real Numbers

[jem05]

hello,
i have 2 gre questions:
1) if i have a sequence x_n in R that goes to 0, then:
a) if f is a continuous function , then f(x_n) is a cauchy sequence. (true ?)
b) if f is a uniformly continuous function , then f(x_n) is a convergent sequence. (true?)

2) lim z--> 0, ^$$\bar{z}$$2 / z² = ?
i got that it does not exist after i took lim r --> 0 ... by converting to exponential form, i
e^-4i \alpha , so it depends on the angle.

thank you...

 

[Dick]

Sure. If f is continuous at x=0 and x_n->0 then f(x_n)->f(0) so the sequence converges. If a sequence converges then it's cauchy. 'Uniformly' doesn't change that. And also, yes, lim $${\bar z}^2/z^2$$ is definitely going to depend on angle.

 

[jem05]

hello, thank you,..
yeah i thought about the " If f is continuous at x=0 and x_n->0 then f(x_n)->f(0) so the sequence converges. " but in wiki, it says continuous functions convert convergent sequences to cauchy ones.
but i would have put in correct even in it said convergent not only cauchy. this is what's confusing me

 

[Dick]

hello, thank you,..
yeah i thought about the " If f is continuous at x=0 and x_n->0 then f(x_n)->f(0) so the sequence converges. " but in wiki, it says continuous functions convert convergent sequences to cauchy ones.
but i would have put in correct even in it said convergent not only cauchy. this is what's confusing me

In the real numbers, 'Cauchy' is the same thing as 'convergent', because the real numbers are 'complete'. Try and wiki that one and see if you get it. In the rationals a sequence may be 'Cauchy' but not 'convergent' because the limit point might not exist, because it might be irrational. In general, 'convergent' is stronger than 'cauchy'. If it's convergent it's cauchy, but not necessarily vice versa.