What will be the value of 'b'?

Question

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.

Perhaps knowing that the answer is $0$ will point you in the right line of thinking. — Andrew Chin, Oct 19 '20 at 12:40
If $b>0$ then use MVT on $[n, n+1]$ and conclude that $f(n) \to \infty$ as $ n \to \infty$. Similarly rule out $b <0$. — Kavi Rama Murthy, Oct 19 '20 at 12:41
Does this answer your question? What does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f'(x)$? — Souza, Oct 19 '20 at 13:36
Assuming $f(x)=\frac{x}{x+1}$ leads nowhere. You can't prove anything that way. Also, note that $\lim_{x\to\infty} f(x)=1$ is not in itself sufficient to prove that $\lim_{x\to\infty}f'(x)$ exists. (You should try to find a counterexample!) But the question tells you that $\lim_{x\to\infty}f'(x)$ does indeed exist, so you can try assuming that $b\ne 0$ and deriving a contradiction using MVT. — TonyK, Oct 19 '20 at 21:27

score 2 · Answer 1 · answered Oct 19 '20 at 12:50

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 $b<0$.)

score 2 · Answer 2 · edited Oct 19 '20 at 13:33

2

$\lim\limits_{x \to \infty} f(x)=1$ 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 $\lim\limits_{x \to \infty}f'(x)=0$.

edited Oct 19 '20 at 13:33

Angelo

13,952

answered Oct 19 '20 at 12:57

Anwesha1729

352

I am looking for a formal method. but this is easy(+1). – Oct 19 '20 at 13:02
1

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. – Anwesha1729 Oct 19 '20 at 13:10
1

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. – TonyK Oct 19 '20 at 21:35
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. – Anwesha1729 Oct 20 '20 at 11:10
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. – TonyK Oct 20 '20 at 14:19

What will be the value of 'b'?

2 Answers2