Suppose f: D--->R and g: D--->, x_0 is an accumulation point of D and f and g have limits at x_o. If f(x)<=g(x) for all x elemts of D then lim f(x)<=Lim g(x) where x goes to x_0.
This question and another you posted really have the same idea.
The proofs are very detailed and tedious.
Basically suppose that .
Now use in the definition of limits.
Doing that you can show that a contradiction will quickly follow if you carefully pick the variables.
I will not do this for you.
woah resurfacing an old post lol. But I just came acrossed this exact question in my book. And i'm stuck I was hoping some one could help me/tell me if im going in the right direction. this is what i have so far:
Suppose that lim x-> f(x)=F , lim x-> g(x)=g and suppose that G<F. Choose >0 and let =(F-G/2)>0 . there is delta1>0 such that for 0<|x- |<delta1 x in D then |f(x)-F|<
and there is a delta2>0 such that for 0<|x- |<delta2 then |g(x)-G|<delta2. let delta=max{delta1, delta2}. Now if 0<|x- |<delta, then |(f(x)-g(x))-(F-G)| <=|f(x)-(F-G)+(F-G)-g(x)|<=|f(x)-(F-G)|+|(F-G)-g(x)|=|f(x)-(F-G)+(F-G)-g(x)|<(F-G)/4+(F-G)/4=(F-G)/2= .
I'm not sure but am I on the right track with this? If not where am I wrong?
oops i did that completely wrong. Heres what I have now:
Suppose lim_x-> f(x) >lim_x-> g(x). By the simple limit theorem, lim_x-> [f(x)-g(x)]>0. Hence we know there exists >0, delta0>0 such that if for all x with 0<|x- |<delta0, we have f(x)-g(x)> . Therefore for all x with 0<|x- |<delta0, f(x)>=g(x) which is a contradiction.
I love how you clearly stated that you should get a contradiction and I kept going with my train of thought. I guess that just what you have to do to spark something that works. I hope this is finally the right proof lol I went ahead and used the simple limit theorem. As far as my prof, he lets us just state and use what we've already proved in class.