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Let $f(x)$ be a function such that $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to \infty}f'(x)=\beta$. Then (1) $\beta > 1$, (2) $\beta$ must be $0$, (3) $\beta$ need not be $0$ but $|\beta| < 1$, (4) $\beta < -1$.

How I proceed: Take $f(x)=\frac{1}{x}$ then $f'(x)=-\frac{1}{x^2}$. Here $\beta=0$.So,(1) and (4) is not true.I am stuck between (2) and (3). Please help.

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up vote 1 down vote accepted

Firstly, to satisfy the condition $\lim_{x\to\infty}f(x)=1$, your choice of $f$ could be corrected as $f(x)=\frac{1}{x}+1$.

Secondly, by mean value theorem, for every $x$, there exists $c_x\in [x,x+1]$, such that $f'(c_x)=f(x+1)-f(x)$. Then $$\beta=\lim_{x\to\infty}f(c_x)=\lim_{x\to\infty}(f(x+1)-f(x))=0.$$

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you are right.Thank you. – Argha Dec 22 '12 at 11:33
@Argha: You are welcome! – Hu Zhengtang Dec 22 '12 at 11:36

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