Let $f:[1,\infty)\rightarrow \mathbb R $ is a differentiable function which satisfies $$f'(x)=\frac {1}{x^2+(f (x))^2} \text{ and } f(1)=1$$ then find the limit of $f $ as $x \to\infty $

My attempt: So I first thought of making a differential equation and then calculate the limit. But the differential equation formed $y'(y^2+x^2)=1$ is a non standard equation and it cannot be solved. Even calculators on the internet show that "no solution found". However, Wolfram Alpha does provide a graph but not the solution.

Also $f^{\prime\prime}(x)<0$. Can this fact be used in some way?

Next I thought of using Rolle's theorem but I am unable to figure out some way to use it.

Can anyone provide me some idea on how to approach this problem?


2 Answers 2


The inverse function satisfies $$x'(f)=x(f)^2+f^2$$ which is a classical Riccati equation. The pole of its solution will be the limit that you are looking for.

You can find a solution in terms of Bessel functions:

You can try various estimates with methods like in

or you can compute a numerical solution.

  • $\begingroup$ Thank you. I got it! $\endgroup$
    – user185887
    Jun 29, 2018 at 7:44

As $f'\left(x\right) >0, f\left(x\right)$ is an increasing function, so we have, for $t>1,f\left(x\right)>f\left(1\right)=1$, Therefore, for $ t>1$,

$$f'\left(t\right)=1+\int_{1}^xf'\left(t\right) dt$$

$$<1+\int_{1}^x\frac{1}{t^2+1} dt$$

$$<1+\int_{1}^{\infty}\frac{1}{t^2+1} dt$$ $$=1+\frac{\pi}{4}$$;

hence, $\lim_{x\to\infty}f\left(x\right)$ exists and is, at most, $$1+\frac{\pi}{4}$$

  • $\begingroup$ LHS is f(x) and not f'(t). $\endgroup$ Nov 22, 2020 at 7:58

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