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 $1+\dfrac{\pi}{4}.$


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 $\lim\limits_{x \to +\infty}f(x)<1+\dfrac{\pi}{4}$?

  • $\begingroup$ @PeterForeman Sir, $f(x)=-\frac{1}{x^2}<0$ but $\lim\limits_{x \to +\infty}f(x)=0.$ $\endgroup$ – mengdie1982 Apr 22 at 10:42
  • $\begingroup$ Are you asked that the limit is smaller than $1+\pi/2$ or than $1+\pi/4$? $\endgroup$ – kingW3 Apr 22 at 10:43
  • $\begingroup$ @kingW3 No. I just wonder whether the equality with the inequality may hold or not. $\endgroup$ – mengdie1982 Apr 22 at 10:45
  • $\begingroup$ Sorry. A typo in the "problem". Corrected. See the new version. $\endgroup$ – mengdie1982 Apr 22 at 10:48
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    $\begingroup$ Possible duplicate of Limit of function as $x \to\infty $ when $f'(x)$ is given $\endgroup$ – 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 $g(1)=0<g(2)<g(x)$ 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 $$\lim_{x\to\infty}f(x)<\frac\pi4+1$$

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    $\begingroup$ In general, $f(x)>g(x)$ implies $ \lim f(x) \geq \lim g(x)$ not $ \lim f(x) > \lim g(x)$... $\endgroup$ – mengdie1982 Apr 22 at 11:43
  • $\begingroup$ @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 $0<\lim_{x\to\infty} g(2) \leq \lim_{x\to\infty} g(x)$. $\endgroup$ – 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 $$ \lim_{x\to\infty}f(x)\le 1+\frac\pi 4-\delta_M<1+\frac\pi 4.$$


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