If $f$ strictly increasing, analytic on $\mathbb{R}$ and $\lim_{x\to +\infty}f(x)=1$, does it follows that $\lim_{x\to +\infty}f'(x)=0$?

If we drop the assumption that the function is increasing, an easy counterexample is $a(x)=\frac{\sin(x^2)}{x}+1$.

If we drop the analyticity requirement (but keep $C^{\infty}$) a counterexample can be constructed from

$$h(x)=\begin{cases}0&x\le 0\\\exp\left(\frac{-\exp(-1/{(x-1)^2})}{x^2}\right)&x\in (0,1)\\ 1&x\ge 1\end{cases}$$

by setting


It is clear that, if $\lim_{x\to +\infty} f'$ exists, it must be $0$:

In fact, since $0=\lim_{x\to +\infty}\frac{f(x)-1}{x}=\lim_{x\to +\infty}f'(x)$.

Otherwise one can prove it by noting that, since $f'\ge 0$ and $1=\int_0^\infty f'(x)dx$ it is impossible to have $\lim_{x\to +\infty}f'(x)>0$.

However, I do not see how to prove the existence of the limit of how to construct a counterexample (as $b$ is not analytic)

  • $\begingroup$ Are you sure it is wrong ? $\endgroup$
    – EDX
    May 1, 2020 at 13:40

2 Answers 2


$$ f(x) = \int_0^x \exp(-t^4 \sin^2(t))\, dt $$

is a counterexample, after normalizing by a suitable constant so that $ \lim_{x \to \infty} f(x) = 1 $ instead of $ 1.17195 \ldots $

The idea is that the factor $ t^4 $ suppresses the contribution of the integrand at all points too far from the zeroes of $ \sin(t) $ as $ t $ gets large, and $ \sin^2(t) $ is only within $ O(1/N^4) $ of $ 0 $ for $ t $ that's $ O(1/N^2) $ from an integer multiple of $ \pi $. Since $ 1/N^2 $ is a fast enough decaying sequence, the integral converges. It's obviously an analytic function.

Edit: Here is a more formal argument. Consider the integral

$$ \int_{N \pi}^{(N+1) \pi} \exp(-t^4 \sin^2(t)) \, dt \leq \int_0^{\pi} \exp(-\pi^4 N^4 \sin^2(t))$$

Split the interval $ [0, \pi] $ into two pieces, $ [1/N^{5/4}, \pi - 1/N^{5/4}] $ and its complement. Over this interval,

$$ \sin^2(t) \geq \sin^2(1/N^{5/4}) = 1/N^{5/2} + O(1/N^5) $$

so that $ \exp(-\pi^4 N^4 \sin^2(t)) \leq \exp(-\pi^4 N^{3/2} + O(N^{-1})) $, which is obviously summable by comparison to the geometric series, say. So the contribution to the integral from this part is $ L^1 $. The other part of the integral is over a set of measure $ 2/N^{5/4} $, and since the integrand is bounded from above by $ 1 $, this part only contributes a term of order $ O(1/N^{5/4}) $, which is also $ L^1 $. We conclude therefore that $ f $ is well-defined by monotonicity and has a finite limit at infinity.

  • $\begingroup$ Why is it an analytic function? $\endgroup$ May 1, 2020 at 13:48
  • $\begingroup$ Just need to be a bit more formal for the conclusion I think, but this is brilliant $\endgroup$
    – EDX
    May 1, 2020 at 13:48
  • $\begingroup$ So in this case, the limit $f'(x)$ does not exist. Suppose the original post had made the assumption that $f'(x)$ existed. $\endgroup$ May 1, 2020 at 13:48
  • $\begingroup$ @ClementYung because the integrand is analytic. $\endgroup$
    – EDX
    May 1, 2020 at 13:49
  • $\begingroup$ @user2379888 They didn't actually ;) $\endgroup$
    – EDX
    May 1, 2020 at 13:49

I was thinking we could write the limit as $$\lim_{x \to \infty}\frac{e^xf(x)}{e^x}=1$$ Since this is an indeterminate form, we can apply L'Hopital's Rule, $$\therefore \lim_{x \to \infty}\frac{e^xf(x)}{e^x}=\lim_{x \to \infty}\frac{e^xf(x)+e^xf'(x)}{e^x}=1$$ $$\therefore \lim_{x \to \infty}(f(x)+f'(x))=1 \implies \lim_{x \to \infty}f'(x) = 0$$

If I have made some error, please feel free to correct me.

  • $\begingroup$ L'hopital works only if the limit of the derivative exists, which is the hard part to prove here $\endgroup$
    – user515010
    May 1, 2020 at 13:47
  • 1
    $\begingroup$ Ahhh, I missed that. I thought when you said that the function is analytic, it will have a derivative and its limit will exist :) $\endgroup$ May 1, 2020 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy