Proving "f(n) & s Have Same Sign as n Approaches Infinity

[cornstarch]

Homework Statement
if lim f(n) = s , prove that lim (f(n))^{1/3} = s^{1/3}. How do you know that as n approaches infinity, f(n) and s have the same sign. n is just an index in this case and f is not a function but a sequence.

The attempt at a solution
so we know that |f(n) - s| < epsilon, using a^3 - b^3 I can factor:
|f(n)^{1/3} - s^{1/3}|*|f(n)^{2/3} + (f(n)*s)^{1/3} + s^{1/3}| < epsilon. Divide both sides and we get |f(n)^{1/3} - s^{1/3}| < epsilon/(|f(n)^{2/3} + (f(n)*s)^{1/3} + s^{1/3}|) which proves the first part. I don't know how to argue that as n approaches infinity, f(n) and s have the same sign.

 

[vela]

You still have work to do to finish the proof of the first part.

You should have no problems showing f(n) and s have the same sign if you understand the definition of the limit. Suppose s>0. You can find an interval around s that only contains positive numbers, right? So...

 

[cornstarch]

what am i missing from the first part of the proof?

 

[vela]

To prove a limit using the definition, you assume ε>0, and then show that for the appropriate choice of N, n>N implies that |f(n)-L|<ε. So first, you need to find N.