Prove or disprove.

Let $f$ be a real-valued function defined on $[1, \infty)$, satisfying $f(1) = 1$ and $f'(x) = \frac{1}{x^2 + [f(x)]^2}$. Then $\lim\limits_{x \rightarrow \infty} f(x)$ exists

I honestly have no idea where to begin. When I rearranged the terms for $f'(x)$, it kinda looked like the mean value theorem. Any help would be much appreciated.


2 Answers 2


$f(x) =1+\int_1^{x} \frac 1 {t^{2}+(f(t))^{2}}\, dt \leq 1+\int_1^{x} \frac 1 {t^{2}}\, dt=2-\frac 1 x<2$ for $x>1$ . Since $f' \geq 0$, $f$ is increasing. Hence $\lim_{x \to \infty} f(x)$ exists. As requested I am providing a proof without integration: by MVT there exists $t \in (n,n+1)$ such that $f(n+1)-f(n)= \frac 1 {t^{2}+f(t)^{2}} \leq \frac 1 {t^{2}}\leq \frac 1 {n^{2}}$. You can easily use this to to see that $\{f(n)\}$ is bounded. Since $f$ is increasing it follows that $f$ is bounded on $[1,\infty)$.

  • $\begingroup$ @user439126 I have added a second proof that avoids integration. $\endgroup$ Jan 31, 2019 at 8:04
  • $\begingroup$ Thank you for the help. $\endgroup$
    – user439126
    Jan 31, 2019 at 10:30
  • $\begingroup$ Can you elaborate more on how its follows that $f$ is bounded from $f$ increasing? $\endgroup$
    – user439126
    Jan 31, 2019 at 11:00
  • $\begingroup$ @user439126 If $x \in (1,\infty)$ there exists an integer $n$ such that $n \leq x \leq n+1$. If $f(n) \leq M$ for all $n$ then we get $f(x) \leq f(n+1) \leq M$. $\endgroup$ Jan 31, 2019 at 11:40
  • $\begingroup$ Unfortunately, I'm having trouble seeing how one can show easily that $\{f(n)\}$ is bounded from $f(n +1) - f(n) \le \frac{1}{n^2}$. I know I can bound $\frac{1}{n^2}$ since the related sequence converges to 0, but I can't seem to see what to do on with LHS. $\endgroup$
    – user439126
    Jan 31, 2019 at 15:34

Hint: show that $f$ is

  • monotonous
  • bounded

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