2

$f$ is a differentiable function on real line such that $$\lim_{x \to \infty}f(x)=1$$ and $$\lim_{x \to \infty}f'(x)=\alpha$$. Then what can be said about $\alpha$

  1. $\alpha=0$
  2. $\alpha$ may not be $0$ but $|\alpha| \le1$
  3. $\alpha\geq1$
  4. $\alpha\leq-1$

I cannot think of any viable option other than 1,but i cannot prove it. Any hint or solution would be appreciated!

Community
  • 1
Prof. Shanku
  • 851
  • Surely for a limit to exist at infinity the gradient as it tends to that point must tend to zero – Henry Lee Jul 18 '19 at 12:51
  • @HenryLee intuitionally it seems correct but how can i prove it? – Prof. Shanku Jul 18 '19 at 12:54
  • 1
    try applying the mean-value theorem – peek-a-boo Jul 18 '19 at 12:58
  • 2
    Use the mean value theorem. For any a and b, there exist c between a and b such that $f(b)- f(a)= f'(c)(b- a)$. Take the limit as b goes to infinity. – user247327 Jul 18 '19 at 12:59
  • 1
    https://math.stackexchange.com/questions/1773890/limit-of-the-derivative-of-a-function – Ruben Jul 18 '19 at 13:05
  • Maybe so? $\lim\limits_{x\rightarrow\infty}f'(x)=\lim\limits_{x\rightarrow\infty}\lim\limits_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\rightarrow0}\lim\limits_{x\rightarrow\infty}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\rightarrow0}\frac{0}{h}=0.$ – Michael Rozenberg Jul 18 '19 at 13:06
  • 1
    @MichaelRozenberg Are you sure you can swap those limits? I had the same solution but was not sure about that step. – Ruben Jul 18 '19 at 13:10
  • @HenryLee Yes, if the limit of the derivative exists it must be $0$. In fact the limit need not exist. – David C. Ullrich Jul 18 '19 at 14:07
  • 1
    @MichaelRozenberg You need some argument (which I doubt exists) to justify swapping those limits. (If limits commuted this way just automatically then a lot of theorems would be trivial, and a lot of non-theorems would be true...) – David C. Ullrich Jul 18 '19 at 14:10
  • @David C. Ullrich It's given that the limit exists. See please better the given. I really don't like your tone. – Michael Rozenberg Jul 18 '19 at 14:21
  • @MichaelRozenberg What "tone"? What you said is wrong, or at least contains a huge gap. So I said so. It's a simple fact that many theorems in analysis are of the form $\lim_x\lim_y=\lim_y\lim_x$; mentioning this is the easiest way to show that you really can't argue this way (without additional justification). – David C. Ullrich Jul 18 '19 at 14:31
  • 1
    @MichaelRozenberg Fine, the limit exists. How does that show that $\lim_{x\to\infty}\lim_{h\to0}=\lim_{h\to0}\lim_{x\to\infty}$? (If $f(n,m)=\frac n{n+m}$ then $\lim_{n\to\infty}\lim_{m\to\infty}$ and $\lim_{m\to\infty}\lim_{n\to\infty}$ both exist, but they're not equal.) – David C. Ullrich Jul 18 '19 at 14:35
  • @MichaelRozenberg Above is an example where you can't swap limits. An example of an (easy) theorem that would be trivialized if swapping limits always worked: If $f_n\to f$ uniformly and each $f_n$ is continuous then $f$ is continuous. Just stating "$\lim_{x\to a}\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}\lim_{x\to a}f_n(x)$" does not prove that. (Because if that argument were correct it could be used to show that a pointwise limit of continuous functions is continuous...) – David C. Ullrich Jul 18 '19 at 14:47

3 Answers3

2

By the process of elimination:

  • Taking $f(x)=1$ eliminates answers $3$ and $4$.
  • Let's assume that $\alpha$ could be nonzero, that is, there exists some $f$ that satisfies both conditions, and $\alpha\neq 0$. Let $\lambda>\neq0$. Then, define $$g(x)=\lambda\cdot f(x) - \lambda + 1.$$ It is then easy to see that $g$ satisfies condition $1$, i.e. it has a limit of $1$, and its derivative has a limit of $\lambda$. Therefore, we have just proven that if some nonzero $\alpha$ is possible, then all real values are possible, so $|\alpha|\leq 1$ is not the correct answer either.

Therefore, $\alpha=0$ must be the correct answer.

Now I know, that's not what we really want here. We'd rather actually prove that $\alpha=0$. To do that, consider the following facts:

  • If $\alpha > 0$, then there exists some $M$ such that $f'(x)>\frac\alpha2$ for all $x>M$.
  • $f(x)=\int_0^x f'(t) dt + f(0)$
  • For any integrable $g$, we have $\int_0^x g(t) dt=\int_0^M g(t)dt + \int_M^x g(t) dt$.
  • For any pair of integrable functions $f, g$, if $g(x)>h(x)$ for all $x\in [a,b]$ then $\int_a^b g(x)dx > \int_a^b h(x)dx$.
5xum
  • 123,496
  • 6
  • 128
  • 204
0

For sufficiently large $ x $ , $ f'$ is locally integrable so we can apply FTC . Let $ a> 0$, By FTC $\lim_{x \to \infty} \int_{x-a}^{x+a} f'(y) dy=\lim_{x \to \infty}(f (x+a)-f (x-a))=0$

If you take $|\alpha| >0$ you get contradiction

ibnAbu
  • 1,966
0

Because $\lim_{x \to \infty}f(x)=1$, $y=1$ is a horizontal asymptote of the graph of $f(x)$.

Therefore, $\lim_{x \to \infty}f'(x)=0$.

DDS
  • 3,199
  • 1
  • 8
  • 31