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I am looking for a function $f$ that is differentiable and $f'(x) \ge c \gt 0$ for all $x \in \mathbb{R}$ and $\lim\limits_{x\to\infty}f(x) \ne \infty$?

Is there such function, or am I wasting my time? Thanks!

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What does the mean value theorem tell you about such functions? –  dls May 23 '12 at 6:10
    
@AmihaiZivan: Your example doesn't work since $\frac{1}{1+x^{2}}\leq 1$ for all $x\in\mathbb{R}$ (instead of $\geq 1$!). As Arturo Magidin demonstrated below, a function with such properties does not exist. –  Thomas E. May 23 '12 at 8:13
    
yup, silly mistake. note to self: always take 1-2 minutes to review your post.. –  Amihai Zivan May 23 '12 at 10:31

2 Answers 2

up vote 5 down vote accepted

Pick an $a\gt 0$. Then by the Mean Value Theorem, there exists a point $r$, $a\lt r\lt a+M$, such that $f'(r)=\frac{1}{M}(f(a+M)-f(a))$. That means that $$f(a+M) - f(a) = Mf'(r) \geq Mc$$ hence for every $M\gt 0$ we have $$f(a+M) \geq f(a)+Mc.$$

What happens as $M\to\infty$?

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If $f'(x)>c$ for every $x$, then $f(x)-f(0) > c x$ for every $x>0$. Hence $$\liminf_{x \to +\infty} f(x) \geq \liminf_{x \to +\infty} \left(f(0)+cx \right) = +\infty.$$

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