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Let $f\colon (0,\infty) \to\Bbb R$ be differentiable and let $A$ and $B$ be real numbers. Prove that if $f(t) \to A$ and $f′(t) \to B$ as $t \to \infty$ then $B = 0$.

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By the Mean Value Theorem, for every positive integer $n$, $$\frac{f(n+1)-f(n)}{1}=f'(c_n),$$ for some $c_n$ between $n$ and $n+1$. Since $f(n+1)-f(n)\to 0$, $f'(c_n)$ must have small absolute value when $n$ is large. So if the limit of $f'(x)$ exists, it must be $0$.

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Assume $B>0$ (the case $B<0$ is symmetric to this). Then there exists $t_0$ such that $f'(t)>\frac12 B$ for $t>t_0$. Then for $t>t_0$ we have $f(t)-f(t_0)=f'(\xi)(t-t_0)>\frac B2 (t-t_0)$ and hence $f(t)\to +\infty$ as $t\to +\infty$.

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