Let $f:[1, \infty ) \rightarrow R$ and $C^1([1, \infty ))$ such that

$f'(x)=\frac{1}{x^2 +\sin^2x+f(x)}$

Prove $\lim\limits_{x \to \infty }f'(x)=0$. How to check that the limit $\lim\limits_{ x \to \infty }f(x)$ exists?


You don't need the existence of that limit to prove $\lim_{x\rightarrow\infty} f'(x)=0$. The fact that $f$ is bounded is enough for the proof.

Let $K=\sup_{[1,\infty)} |f(x)|$. Then, for $x$ sufficiently big, it holds $$\frac{1}{x^2+1+K}\leq f'(x)\leq\frac{1}{x^2-K}.$$

Now, by sending $x\rightarrow\infty$ we get $\lim_{x\rightarrow\infty} f'(x)=0$.

  • 4
    $\begingroup$ How do you now that $f(x)$ is bounded? $\endgroup$ – lisyarus Apr 23 '15 at 14:02
  • $\begingroup$ I assumed that $C^1([1, \infty))$ means continuous, with continuous derivative, both bounded. But, as I looked on the internet now it doesn't always have to be like that. Of course, it doesn't work this way if it means just continuous plus continuous derivative. $\endgroup$ – katarina Apr 23 '15 at 14:13

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