# Problem

Let $$f(x)$$ satisfy that $$f(1)=1$$ and $$f'(x)=\dfrac{1}{x^2+f^2(x)}$$. Prove that $$\lim\limits_{x \to +\infty}f(x)$$ exists and is less than $$1+\dfrac{\pi}{4}.$$

# Proof

Since $$f'(x)=\dfrac{1}{x^2+f'(x)}>0$$, $$f(x)$$ is strictly increasing. Thus, $$f(x)>f(1)=1$$ holds for all $$x>1$$, and $$\lim\limits_{x \to +\infty}f(x)$$ equals either the positive infinity or some finite value.

Notice that, $$\forall x>1:$$ \begin{align*} f(x)-f(1)&=\int_1^x f'(t){\rm d}t=\int_1^x \frac{1}{t^2+f^2(t)}{\rm d}t<\int_1^x\frac{1}{t^2+1}{\rm d}t=\arctan x-\frac{\pi}{4}. \end{align*} Therefore $$f(x)<\arctan x-\frac{\pi}{4}+1<\frac{\pi}{2}-\frac{\pi}{4}+1=1+\frac{\pi}{4},$$ which implies that $$f(x)$$ is bounded upward. Thus,$$\lim\limits_{x \to +\infty}f(x)$$ exists. Take the limits as $$x \to +\infty$$, we have $$\lim\limits_{x \to +\infty}f(x)\leq 1+\dfrac{\pi}{4}.$$ Can we cancel the equality mark here? In another word, can we obtain $$\lim\limits_{x \to +\infty}f(x)<1+\dfrac{\pi}{4}$$?

• @PeterForeman Sir, $f(x)=-\frac{1}{x^2}<0$ but $\lim\limits_{x \to +\infty}f(x)=0.$ – mengdie1982 Apr 22 at 10:42
• Are you asked that the limit is smaller than $1+\pi/2$ or than $1+\pi/4$? – kingW3 Apr 22 at 10:43
• @kingW3 No. I just wonder whether the equality with the inequality may hold or not. – mengdie1982 Apr 22 at 10:45
• Sorry. A typo in the "problem". Corrected. See the new version. – mengdie1982 Apr 22 at 10:48
• Possible duplicate of Limit of function as $x \to\infty$ when $f'(x)$ is given – LutzL Apr 22 at 15:26

The function $$g(x)=\int_1^x\frac{1}{t^2+1}{\rm d}t-\int_1^x \frac{1}{t^2+f^2(t)}{\rm d}t$$ Is strictly increasing and $$g(1)=0 for $$x>2$$ hence $$\lim_{x\to\infty}g(x)\geq g(2)>0$$ so$$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}(\frac\pi4-(f(x)-1))=\lim_{x\to\infty}(\frac\pi4+1-f(x))>0$$
So $$\lim_{x\to\infty}f(x)<\frac\pi4+1$$
• In general, $f(x)>g(x)$ implies $\lim f(x) \geq \lim g(x)$ not $\lim f(x) > \lim g(x)$... – mengdie1982 Apr 22 at 11:43
• @mengdie1982 Yeah but I didn't use that in my proof. I've used that $g$ is strictly increasing and that $g(x)>0$ for $x>1$. Using that you could prove $g(2)<g(x)$ for $x>2$ hence $0<\lim_{x\to\infty} g(2) \leq \lim_{x\to\infty} g(x)$. – kingW3 Apr 22 at 11:51
Fix $$M>1$$ and for $$x>M$$ break up your estimate as $$f(x)-f(M)<\arctan x- \arctan M$$ and $$f(M)-f(1)<\arctan M-\frac\pi 4,$$ so there is a positive constant (i.e., depending only on $$M$$, but not on $$x$$) $$\delta_M:=\arctan M-\frac\pi 4-f(M)+f(1)$$. Then for $$x>M$$, $$f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<\arctan x-\frac \pi 4-\delta_M$$ and so $$\lim_{x\to\infty}f(x)\le 1+\frac\pi 4-\delta_M<1+\frac\pi 4.$$