# Prove that if $f'(x)=\frac{1}{x^2+[f(x)]^2},\forall x\in[1,+ \infty[$ then $\lim_{x\to+\infty} f(x)< 1+\frac{\pi}{4}$

Let $f:[1,+ \infty[ \rightarrow \mathbb{R}$, differentiable , $f(1)=1$ and

$$f'(x)=\frac{1}{x^2+[f(x)]^2},\forall x\in[1,+ \infty[$$

Prove that, $$\lim_{x\to+\infty} f(x)< 1+\frac{\pi}{4}$$

I managed to prove $\lim_{x\to+\infty} f(x)\leqslant2$, but I can't get the above result. Any help would be greatly appreciated.

• Do you really require the strict inequality? – David Mitra Jul 1 '14 at 21:58
• Yes, the question I got is exactly has stated. – King Ghidorah Jul 1 '14 at 21:59

Clearly the derivative is bounded by $1/(x^2+1)$, so using the FTC
$$f(x)=1+\int_1^x{dt\over 1+f(t)^2}\le 1+\int_1^x{dt\over 1+t^2}\stackrel{x\to\infty}{\longrightarrow}1+{\pi\over 4}.$$
• This argument won't give a strict inequality; but it's easily fixed ($\int_1^2 {1\over x^2+(f(x))^2}\,dx<\alpha+\int_1^2{1\over 1+x^2}\,dx$ for some $\alpha>0$ ). (And, I think you want an "$x^2$" downstairs in your first integral.) – David Mitra Jul 1 '14 at 22:16
• Yes, the last $\epsilon$-bit of the proof is that. Thanks David for covering for my disinclination towards typing. :-) – Adam Hughes Jul 1 '14 at 22:21