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Suppose $f: \mathbb R \rightarrow \mathbb R$ is a function with linear growth and $\lim_{x\rightarrow \infty} f'(x)$ exists and is finite. I'm trying to show that $\lim_{x\rightarrow \infty} f(x)/x$ also exists and establish conditions under which it is equal to $\lim_{x\rightarrow \infty} f'(x)$.

I found the answer to this here: If $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x \to \infty} \frac {f(x)}{x} = L$

Jong
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  • What have you tried? Where are you getting stuck? – Bonnaduck Nov 07 '21 at 22:16
  • Suppose the limit inferior is not equal to the limit superior of $f(x)/x$. Then at $x$ such that $f(x)/x$ is equal to its limit superior, intuitively I think it must also be equal to $f'(x)$. Similarly for $x$ such that $f(x)/x$ is equal to its limit inferior. But then if $f'(x)$ converges, this establishes that the limit inferior and superior of $f(x)/x$ must be equal. But I am having a bit of trouble showing formally that at $x$ such that $f(x)/x$ is equal to its limit superior, it is also equal to $f'(x)$. – Jong Nov 07 '21 at 22:19
  • So I think this immediately follows from the characterization of a local maximum or minimum. In other words, from the FOC for $\max_x f(x)/x$, $f'(x)=f(x)/x$. So then if $f'(x)$ converges, it must be that $\max_x f(x)/x$ and $\min_x f(x)/x$ become arbitrarily close to $f'(x)$ and therefore, lim sup = lim inf. – Jong Nov 07 '21 at 22:31
  • You might get better attention to the question by editing those comments into the body of the question where people will see them. – David K Nov 08 '21 at 03:10
  • What is your definition of "linear growth"? To me that term means precisely that $\lim_{x\to\infty} \frac{f(x)}x$ exists.... – Greg Martin Nov 08 '21 at 05:20
  • The definition of linear growth is there exists an $M$ such that $|f(x)| \leq M|x|$ for sufficiently large $x$. $f(x) could still oscillate, in which case the limit doesn't exist. – Jong Nov 09 '21 at 04:23

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