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Suppose f(x) <= g(x). Prove that lim [x->a] f(x) <= lim[x->a] g(x).

I've been able to prove it by contradiction. I let lim [x->a] f(x) = L and lim[x->a] g(x) = M and I suppose that L > M. I then went on to choose epsilon = (L - M)/2 and a contradiction easily follows. But my problem is how do you arrive/choose this choice of epsilon? I just immediately thought of using this epsilon, but if I hadn't I don't know how/why I would have arrived at this epsilon. Can someone please help me out with this?

Thanks

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# Limit Proof

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