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

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.

$$f(x) =1+\int_1^{x} \frac 1 {t^{2}+(f(t))^{2}}\, dt \leq 1+\int_1^{x} \frac 1 {t^{2}}\, dt=2-\frac 1 x<2$$ 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 $$[1,\infty)$$.
• Can you elaborate more on how its follows that $f$ is bounded from $f$ increasing? – user439126 Jan 31 '19 at 11:00
• @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 $f(x) \leq f(n+1) \leq M$. – Kavi Rama Murthy Jan 31 '19 at 11:40
• 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. – user439126 Jan 31 '19 at 15:34
Hint: show that $$f$$ is