Suppose $f\in C^1[0,+\infty) $ ,$f(0)=1$ ,$f'(x)=\frac{1}{x^2+f(x)} $ Suppose $f\in C^1[0,+\infty) $ ,$f(0)=1$ ,$f'(x)=\frac{1}{x^2+f(x)} $
Proof


*

*$\lim_{x\to +\infty }f(x)$ exsits

*$\lim_{x\to +\infty }f(x)\le1+\frac{\pi}{2} $
All I attempted is connected it to Cauchy Criterion since we just need to proof it exists, but I have no clue for this. 
 A: First, $f'(x)>0$ and hence $f(x)$ is increasing in $[0,\infty)$. Second
$$ f'(x)=\frac{1}{x^2+f(x)}\le\frac{1}{x^2+f(0)}=\frac{1}{x^2+1} $$
and so
$$ f(x)-f(0)\le \int_0^x\frac{1}{t^2+1}dt\le\int_0^\infty\frac{1}{t^2+1}dt=\frac{\pi}{2} $$
Namely $f(x)\le1+\frac{\pi}{2}$ or $f(x)$ is bounded in $[0,\infty)$ and hence
$\lim_{x\to\infty}f(x)$ exists.
A: Sketch: Suppose you can show $f>0$ everywhere; I'll leave this part to you. Then $f'>0,$ which implies $f$ is strictly increasing. Thus
$$f(x)-f(0) = \int_0^x f'(t)\,dt = \int_0^x \frac{1}{t^2 +f(t)}\,dt \le \int_0^x \frac{1}{t^2 +1}\,dt = \arctan x\le \frac{\pi}{2}.$$
This implies $f$ is bounded above, and since $f$ is strictly increasing, it has a limit at $\infty.$ The bound on the limit follows easily.
A: You can show that $f$ has non-negative derivative. Then  $f(0)=1$ implies that $f$ is nondecreasing and $f\ge 1$. Hence 2 implies 1. To prove 2 compute $$
|f(x)|=|1+\int_0^x \frac1{y^2+f(y)}\,dy|\le 1+\int_0^\infty \frac1{y^2+1}\,dy=1+\pi/2.
$$
