Show that $\lim_{x\rightarrow\infty}f(x)$ exists where $f'(x) = \frac{1}{x^2+(f(x))^2}$

Question

Let $f(x)$ be a real-valued differentiable function on $[1,\infty)$ satisfying $f(1)=2$ and

$f'(x) = \frac{1}{x^2+(f(x))^2}$.

Show that $\lim_{x\rightarrow\infty}f(x)$ exists and that it is less than $2+\frac{\pi}{4}$.

My attempt: It can be seen fairly easily that the derivative lies in the interval $(0,\frac{1}{5}]$ for $x\in[1,\infty)$. My first thought was to try and bound the derivative and then integrate, i.e. take

$\int_{1}^{N}f'(x) = f(N)-f(1)\leq\int_1^N\frac{1}{5} = \frac{N}{5}-\frac{1}{5},$ but this bound didn't lead me anywhere.

Any other ideas? Is this the right direction?

Any help appreciated!

Answer 1

It is clear that $f'(x)>0$ for all $x\ge1$. This implies that $f$ is increasing, that $f(x)\ge2$ for all $x\ge1$ and that $$ 0\le f'(x)\le\frac{1}{x^2+4}. $$ If $x>1$ then $$ f(x)=f(1)+\int_1^xf'(t)\,dt\le2+\int_1^x\frac{dt}{t^2+4}=2+\frac12(\arctan\frac{x}{2}-\arctan\frac12)\le2+\frac{\pi}{4}. $$ Since $f$ is increasing and bounded ...