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I have a couple of questions where I use the MVT and I wanted your opinion on the appropriateness of my proofs.

Let $F()$ be a function such that $\lim_{x\to\infty} F'(x)=0 $ and let $G(x,y)$ be a function such that $\lim_{x\to\infty} G(x,y)=\infty $ and $G(x,y_1) > G(x,y_2)$ for all $x, y_1>y_2$. Assume that $F(0)=0$ and that $F$ and $G$ are both always non-negative.

Question: Given a>b, Does $\lim_{x->\infty} F(G(x,a)-F(G(x,b))=0 $?

My Answer: Yes, since by the MVT: $F(G(x,a)-F(G(x,b))=F'(g)[G(x,a)-G(x,b)]$

As $x\to\infty, g\to\infty,$ and thus, $F'(g)\to0$ But, I still have $(G(x,a)-G(x,b))$ and I have to show this is bounded?

Similarly, I have the following question.

Let G be the function as described above and and now assume that $\lim_{x\to\infty} F'(x)=\infty $, would it now follow (by similar reasoning) that $\lim_{x\to\infty} F(G(x,a)-F(G(x,b))=\infty $?

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LaTeX hint: use \to for arrows, like x \to \infty –  Ilya Jan 10 '13 at 17:12
    
Thanks, edited. –  Greg Jan 10 '13 at 19:50

1 Answer 1

up vote 1 down vote accepted

Yes, you would need an extra assumption to guarantee the boundedness of that difference. Otherwise a counterexample would be given by $F(x) = \sqrt{x}$, $G(x,0) = x^2,\, G(x,1) = 2x^2$.

A similar counterexample can be given for your second question.

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Thanks, I've posted a related follow up question in case youre interested –  Greg Jan 11 '13 at 17:23

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