Problem: $f'(x)>c>0$ for all $x\in [0,\infty)$. Show that $\lim_{x\to \infty}f(x)=\infty.$
My attempt: I took $g(x)=f(x)-cx$ then $g'(x)>0$ clearly. I don't know what to do after this step. Any hints will be much appreciated.
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For any $x>0$, $$f(x) = f(0) + f'(d_x)x$$ for some $d_x$ in between $0$ and $x$. Hence $$f(x) >f(0)+cx$$
Now take limit to infinity.
Siong Thye Goh
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Since $g'(x)>0$, $g$ is increasing. Therefore, for all $x >1$ you have $$g(x) \geq g(1)$$
Thus $$f(x) \geq cx +g(1)$$
N. S.
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we know that $$\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}f'(x)$$from what was given in the question we get $$\lim_{x\to\infty}\frac{f(x)}{x}>c\implies\lim_{x\to\infty}{f(x)}>\lim_{x\to\infty} cx$$and because that for sufficiently large $x$ we have $cx$ diverges hence $f(x)$ also diverges thus $$\lim_{x\to\infty}{f(x)}\to\infty$$
Holo
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I think the answer could be rewrote in a more rigorous way: For sufficiently large $x$, we have $f(x)>cx$. Since $cx$ diverges to $+\infty$, it implies $f$ also diverges to $+\infty$, instead of $$\lim_{x\to\infty}{f(x)}>\lim_{x\to\infty} cx\implies\lim_{x\to\infty}f(x)>\infty\implies\lim_{x\to\infty}f(x)=\infty.$$ – Tianlalu Dec 16 '17 at 18:51
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@WongAustin i edit it, hope it is better this way – Holo Dec 16 '17 at 18:57