# What will be the value of 'b'? [duplicate]

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 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 at 12:41
• @Kavi Rama Murthy: How $f(n)\to \infty$ as $n\to\infty$? – user787636 Oct 19 at 12:46
• @Joseph $f(n+1)>(b/2)+f(n)$ for $n$ large enough. – Kavi Rama Murthy Oct 19 at 13:17
• 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 at 13:36

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 $$b<0$$.)

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

• I am looking for a formal method. but this is easy(+1). – user787636 Oct 19 at 13:02
• 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 at 13:10
• 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 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 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 at 14:19