Does $\lim\limits_{x \rightarrow \infty} f(x)$ exists?

Question

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.

Answer 1

$f(x) =1+\int_1^{x} \frac 1 {t^{2}+(f(t))^{2}}\, dt \leq 1+\int_1^{x} \frac 1 {t^{2}}\, dt=2-\frac 1 x<2$ 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 $[1,\infty)$.

Answer 2

Hint: show that $f$ is

• monotonous
• bounded