1. Continuity proofs

Prove that for continuous f, f(x_n)>=f(y_n) => f(lim x_n)>=f(lim y_n)

I started a proof by contradiction, because that is how I remember doing these types of problems in the past. I said, suppose not, then there is an N such that for n>N, f(y_n)>f(x), but how can I go from there to showing that means there is an n such that f(x_n)>f(y_n). I have a feeling that it is a clever choice of epsilon that I cannot think of.

Thanks!

2. Re: Continuity proofs

Originally Posted by powerhawk
[FONT=Consolas]Prove that for continuous f, f(x_n)>=f(y_n) => f(lim x_n)>=f(lim y_n)
You know that $\displaystyle \lim (f({x_n})) = X\;\& \;\lim (f({y_n})) = Y$ both exist.
Now suppose that $\displaystyle X<Y$. Then let $\displaystyle \varepsilon = \frac{{Y - X}}{2}$.
$\displaystyle (\exists~N)[n\ge N]$ implies $\displaystyle X - \varepsilon < {x_n} < X + \varepsilon\;\; \&$ $\displaystyle Y - \varepsilon < {y_n} < Y + \varepsilon$.