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 $\displaystyle \lim _{x \to x_0 } f(x) = F\,,\,\lim _{x \to x_0 } g(x) = G\,\& \,G < F$.
Now use $\displaystyle \varepsilon = \frac{{F - G}}{2} > 0$ 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.
$\displaystyle \varepsilon=\frac{{F - G}}{2}>0$
iff for each $\displaystyle \varepsilon>0$, there is a delta>0 such that if absvalue(x-$\displaystyle x_0$)<delta, then absvalue(f(x)-F)<$\displaystyle \frac{{F - G}}{2}$.
Similarily, absvalue(g(x)-G)<$\displaystyle \frac{{F - G}}{2}$
I'm not sure what to do with this information?
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-> $\displaystyle x_0$ f(x)=F , lim x->$\displaystyle x_0$ g(x)=g and suppose that G<F. Choose $\displaystyle \epsilon$>0 and let $\displaystyle \epsilon$=(F-G/2)>0 . there is delta1>0 such that for 0<|x-$\displaystyle x_0$|<delta1 x in D then |f(x)-F|<$\displaystyle \epsilon$
and there is a delta2>0 such that for 0<|x-$\displaystyle x_0$|<delta2 then |g(x)-G|<delta2. let delta=max{delta1, delta2}. Now if 0<|x-$\displaystyle x_0$|<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=$\displaystyle \epsilon$.
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->$\displaystyle x_0$ f(x) >lim_x->$\displaystyle x_0$ g(x). By the simple limit theorem, lim_x->$\displaystyle x_0$[f(x)-g(x)]>0. Hence we know there exists $\displaystyle \epsilon0$>0, delta0>0 such that if for all x with 0<|x-$\displaystyle x_0$|<delta0, we have f(x)-g(x)>$\displaystyle \epsilon0$. Therefore for all x with 0<|x-$\displaystyle x_0$|<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.