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Let $f:(0,\infty)\to \Bbb{R}$ be differentiable and $\lim\limits_{x\to \infty}f(x)=1$ and $\lim\limits_{x\to \infty}f'(x)=b$. Find the value of $b$.

I assume $f(x)=\dfrac{x}{x+1}$. I found above is true. But how to prove formally? I wanna prove it by MVT. How to proceed? As given $f$ is differentiable then we can use formula of derivative. But how? Any help will be appreciate.

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  • $\begingroup$ Perhaps knowing that the answer is $0$ will point you in the right line of thinking. $\endgroup$ – Andrew Chin Oct 19 at 12:40
  • $\begingroup$ If $b>0$ then use MVT on $[n, n+1]$ and conclude that $f(n) \to \infty$ as $ n \to \infty$. Similarly rule out $b <0$. $\endgroup$ – Kavi Rama Murthy Oct 19 at 12:41
  • $\begingroup$ @Kavi Rama Murthy: How $f(n)\to \infty$ as $n\to\infty$? $\endgroup$ – user787636 Oct 19 at 12:46
  • $\begingroup$ @Joseph $f(n+1)>(b/2)+f(n)$ for $n$ large enough. $\endgroup$ – Kavi Rama Murthy Oct 19 at 13:17
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    $\begingroup$ Does this answer your question? What does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f'(x)$? $\endgroup$ – Souza Oct 19 at 13:36
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Alternative method: if $f'(x) → b >0$, then by definition, there is some $X>0$ such that

$$x>X \implies |f'(x) - b| > b \implies f'(x) > b/2.$$

Then, for $x > X$,

$$f(x) = f(X) + \int_X^x f'(y)\; dy \geq f(X) + \int_X^x b/2\; dy = f(X) + \frac b2(x-X).$$

I.e., $f(x) \geq \frac b2(x-X).$

(Similarly if $b<0$.)

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$\lim\limits_{x \to \infty} f(x)=1$ or $\lim\limits_{x \to \infty} \dfrac{e^xf(x)}{e^x}=1$ . It is $\dfrac{\infty}{\infty}$ form.

As $f(x)$ is differentible in ($0$, $\infty$), using L’Hospital’s rule, $\lim\limits_{x \to \infty} \dfrac{e^xf(x)}{e^x}= \lim\limits_{x \to \infty} \dfrac{e^xf(x)+ e^xf'(x)}{e^x}= \lim\limits_{x \to \infty} f(x)+ f'(x)$ which is given to be equal to $1$.

Therefore $\lim\limits_{x \to \infty}f'(x)=0$.

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  • $\begingroup$ I am looking for a formal method. but this is easy(+1). $\endgroup$ – user787636 Oct 19 at 13:02
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    $\begingroup$ Intuitively one can imagine that f(x) is approching a fixed value as x tends to infinity so f'(x) must be equal to 0 as x tends to infinity. e^x gives shape to that intuition. $\endgroup$ – Anwesha1729 Oct 19 at 13:10
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    $\begingroup$ This doesn't work. According to your argument, for any differentiable $f$ such that $\lim_{x\to\infty}f(x)=1$, we must have $\lim_{x\to\infty}f'(x)=0$. But this is manifestly false. See for instance this Wikipedia article for the conditions that must be satisfied before L'Hôpital's rule can be invoked. $\endgroup$ – TonyK Oct 19 at 21:35
  • $\begingroup$ It should be mentioned that if limx→∞f′(x) exists, then the limit must be 0. If limx→∞f′(x) exists then this method is correct. $\endgroup$ – Anwesha1729 Oct 20 at 11:10
  • $\begingroup$ Yes. But that is something that you have to prove, using an argument similar to GoodBoy's. Your method is the wrong way to go about this. $\endgroup$ – TonyK Oct 20 at 14:19