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Math Help - Prove lim f(x)<Lim g(x) Thm

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    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 \lim _{x \to x_0 } f(x) = F\,,\,\lim _{x \to x_0 } g(x) = G\,\& \,G < F.
    Now use \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.
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  3. #3
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    Quote Originally Posted by Plato View Post
    This question and another you posted really have the same idea.
    The proofs are very detailed and tedious.
    Basically suppose that \lim _{x \to x_0 } f(x) = F\,,\,\lim _{x \to x_0 } g(x) = G\,\& \,G < F.
    Now use \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.
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  4. #4
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    Quote Originally Posted by Plato View Post
    This question and another you posted really have the same idea.
    The proofs are very detailed and tedious.
    Basically suppose that \lim _{x \to x_0 } f(x) = F\,,\,\lim _{x \to x_0 } g(x) = G\,\& \,G < F.
    Now use \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?
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  5. #5
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    Quote Originally Posted by Plato View Post
    This question and another you posted really have the same idea.
    The proofs are very detailed and tedious.
    Basically suppose that \lim _{x \to x_0 } f(x) = F\,,\,\lim _{x \to x_0 } g(x) = G\,\& \,G < F.
    Now use \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.
    \varepsilon=\frac{{F - G}}{2}>0
    iff for each \varepsilon>0, there is a delta>0 such that if absvalue(x- x_0)<delta, then absvalue(f(x)-F)< \frac{{F - G}}{2}.
    Similarily, absvalue(g(x)-G)< \frac{{F - G}}{2}

    I'm not sure what to do with this information?
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  6. #6
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    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-> x_0 f(x)=F , lim x-> x_0 g(x)=g and suppose that G<F. Choose \epsilon>0 and let \epsilon=(F-G/2)>0 . there is delta1>0 such that for 0<|x- x_0|<delta1 x in D then |f(x)-F|< \epsilon
    and there is a delta2>0 such that for 0<|x- x_0|<delta2 then |g(x)-G|<delta2. let delta=max{delta1, delta2}. Now if 0<|x- 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= \epsilon.

    I'm not sure but am I on the right track with this? If not where am I wrong?
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  7. #7
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    oops i did that completely wrong. Heres what I have now:

    Suppose lim_x-> x_0 f(x) >lim_x-> x_0 g(x). By the simple limit theorem, lim_x-> x_0[f(x)-g(x)]>0. Hence we know there exists \epsilon0>0, delta0>0 such that if for all x with 0<|x- x_0|<delta0, we have f(x)-g(x)> \epsilon0. Therefore for all x with 0<|x- 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.
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  8. #8
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    Quote Originally Posted by alice8675309 View Post
    oops i did that completely wrong. Heres what I have now:

    Suppose lim_x-> x_0 f(x) >lim_x-> x_0 g(x). By the simple limit theorem, lim_x-> x_0[f(x)-g(x)]>0. Hence we know there exists \epsilon0>0, delta0>0 such that if for all x with 0<|x- x_0|<delta0, we have f(x)-g(x)> \epsilon0. Therefore for all x with 0<|x- 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.
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