Differentiable function such that $\lim_\limits{x\to \infty}f(x)=\infty$

Let $f:(1, \infty) \to \mathbb{R}$ be a differentiable function such that $$f'(x)=\frac{x^2-(f(x))^2}{x^2(1+(f(x))^2)}$$ for all $x>1$. Prove that $\lim_\limits{x \to \infty}f(x)=\infty$

From the given relation, I got that $f$ is infinitely differentiable. I tried to get an inequality between $x$ and $f(x)$, and so I tried to prove that the derivative doesn't change signs, so that $f$ is monotonic, but I don't think this works. Also, I noticed that if $f(x)=x$ for a given $x$, then $f'(x)=0$. I also don't see why the domain is given as $(1,\infty)$. Perhaps this is supposed to hint to something.

2 Answers

Claim 1: $f$ is unbounded above.

Proof: By contradiction. Suppose $f(x)\le M$ for all $x\in(1,\infty)$. Then $$(1+f(x)^2)f'(x)=1-\frac{f(x)^2}{x^2}\ge\frac12,\quad x\ge M\sqrt2.$$ Integrating $$f(x)+\frac13\,f(x)^3\ge \frac{x}{2}+C,\quad x\ge M\sqrt2.$$ The function $h(u)=u+u^3/3$ is strictly increasing and has an inverse $h^{-1}$ and $f(x)\ge h^{-1}(x/2+C)$ for large $x$. Since $\lim_{u\to\infty}h^{-1}(u)=+\infty$, we reach a contradiction with the fact that $f$ was assumed bounded.

Claim 2: $f$ has at most one critical point, which is a strict minimum.

Proof: $\xi\in(1,\infty)$ is a critical point of $f$ if and only if $f(\xi)=\xi$. Derivating the equation satisfied by $f$ we get $$f''(\xi)=\frac{(2\xi-2f(\xi)f'(\xi))\xi^2(1+f(\xi)^2)}{\xi^4(1+f(\xi)^2)^2}=\frac{2}{\xi(1+f(\xi)^2)}>0.$$ If $f$ had two critical points, both would be minimums, and there would be a maximum between them, an impossibility.

Finally, from Claims 2 it follows that $f$ is eventually increasing, and since it is unbounded by Claim 1, we must have $\lim_{x\to\infty}f(x)=+\infty$.

• the inequalities should be other way around I suppose. then it still doesn't prove it. – Arian Mar 28 '18 at 14:39
• Oh my! I do not know what I was thinking. – Julián Aguirre Mar 28 '18 at 14:51
• I think I have salvaged the argument. – Julián Aguirre Mar 28 '18 at 15:36

If $f(x)=0$ then $f'(x)=1$ and if $f'(x)=0$, then $f''(x)>0$. So $f$ has at most one local minimum, and either $f'$ is always negative (then so is $f$, and $f$ tends to $-\infty$ faster than $-x$) or $f'$ changes sign once from negative to positive or $f'$ is always positive. So $f$ is increasing on $(a,\infty)$ for some $a\ge 1$. Either $f$ tends to $\infty$ or $f$ is bounded. However, if $f$ is bounded, $f'(x)\to 1$ as $x\to\infty$, a contradiction.