how to compute this limits given these conditions. 
if $f(1)=1$ and $f'(x)=\frac{1}{x^2+[f(x)]^2}$ then compute $\lim\limits_{x\to+\infty}f(x)$

i tried to write it was
$$\frac{dy}{dx}=\frac{1}{x^2+y^2}\\
(x^2+y^2)\frac{dy}{dx}=1\\
(x^2+y^2)dy=dx$$
by the help
$$\begin{align}
f(x)&\le1+\int_1^x\frac{dt}{1+t^2}\\
&\le1+\arctan t\bigg|_1^x\\
&\le1+\arctan x-\arctan 1\\
&\le1+\arctan x-\frac{\pi}{4}
\end{align}$$
so
$$\begin{align}
\lim\limits_{x\to+\infty}f(x)&\le\lim\limits_{x\to+\infty}1+\arctan x-\frac{\pi}{4}\\
&\le1+\frac{\pi}{2}-\frac{\pi}{4}\\
&\le1+\frac{\pi}{4}=\frac{4+\pi}{4}
\end{align}$$

From the other answer it is clear that $f(x)$ is strictly increasing in $[1,\infty)$ and is bounded above by $A=1+\dfrac{\pi}{4}$ and hence $\lim\limits_{x\to\infty}f(x)=L$ exists and we have $1<L\leq A$ and $f(x)< L\leq A$. Next we can see that
\begin{align}f'(x)=\frac{1}{x^{2}+\{f(x)\}^{2}}&>\frac{1}{x^{2}+L^{2}}\notag\\
\Rightarrow \int_{1}^{x}f'(t)\,dt&>\int_{1}^{x}\frac{dt}{t^{2}+L^{2}}\notag\\
\Rightarrow f(x)&>1+\frac{1}{L}\tan^{-1}\left(\frac{x}{L}\right)-\frac{1}{L}\tan^{-1}\left(\frac{1}{L}\right)\notag\\
\Rightarrow L&\geq 1+\frac{\pi}{2L}-\frac{1}{L}\tan^{-1}\left(\frac{1}{L}\right)=1+\frac{\tan^{-1}L}{L}\notag\end{align}
Hence we need to find the least number $L \in(1,A]$ which satisfies the above inequality. From online graphing calculator it looks like the desired value of $L$ is around $1.6268\dots$
Another take on this problem is as follows. Let $g(x)=g_{1}(x)=1+\dfrac{\arctan x}{x}$ and let $g_{n}(x)$ denote $n^{\text{th}}$ iterate of $g(x)$. Then $A=g_{1}(1)$ and we can easily show that $$g_{2n}(1)\leq L\leq g_{2n+1}(1)$$ for all $n$. It follows that $L$ is given as the unique root of $g(x)=x$ which is around $1.6268\dots$
