# Thread: Prove lim f(x)<Lim g(x) Thm

1. ## Prove lim f(x)<Lim g(x) Thm

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.

2. Originally Posted by kathrynmath
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.

3. Originally Posted by Plato
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.
Ok, I'll see what I can do with that.

4. Originally Posted by Plato
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.
So just using this in the definition of limits should be all I need to do, right?

5. Originally Posted by Plato
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?

6. 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?

7. 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.

8. Originally Posted by alice8675309
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.
You get:

|f(x) - F| < (F - G)/2 and |g(x) - G| < (F - G)/2

Expand both of these inequalities and "solve" for f(x) and g(x) and you'll see that you'll be able to "connect" the two which will yield a contradiction.