Can f have a finite limit at infinity? The function $ f:\mathbb R\rightarrow \mathbb R$ is differentiable such that $f(0)=0$ and $(1+x^2)f'(x)\geq{1+(f(x))^2}$ for every $ x\in \mathbb R$ . Can $f$ have a finite limit at infinity?
 A: Consider the Cauchy's problem $\left(1+x^2\right)y'=1+y^2$ with initial value $y\left(x=0\right)=0$. Then we have 
$$\frac{y'}{1+y^2}=\frac{1}{1+x^2}$$
whence after integration
$$y\left(x\right)=x.$$
Now we prove that $f\geq y$.
Let $I=\{x\in\mathbb{R}\mid f\left(x\right)\geq y\left(x\right)\}$.
$I\neq \emptyset$ because $0\in I$. As $f$ and $y$ are differentiable, they are continuous and if $\left(x_n\right)$ is a convergent sequence in $I$, then $f\left(x\right)=\lim f\left(x_n\right)\geq\lim y\left(x_n\right)= y\left(x\right)$ so $I$ is closed.
Now, for all $\varepsilon>0$ there is $0<\eta<\varepsilon$ such that
$$f\left(\varepsilon\right)
=f\left(\varepsilon\right)-f\left(0\right)
=f'\left(\eta\right)\left(\varepsilon-0\right)
=f'\left(\eta\right)\varepsilon
=f'\left(\eta\right)y\left(\varepsilon\right)$$ and then
$$f\left(\varepsilon\right)
=\frac{1+f\left(\eta\right)^2}{1+\eta^2}y\left(\varepsilon\right)
\geq y\left(\varepsilon\right)$$
since $f\left(\eta\right)\geq\eta$ (because $f'\left(0\right)\geq1$ and hence $f$ is increasing). Do the same for $\varepsilon<0$. Then $I$ is open.
Since $\mathbb{R}$ is connected, $I=\mathbb{R}$ and $f\geq y$.
Finally, as $\lim_{x\rightarrow+\infty}x=+\infty$ and as $f\geq y$, we can say that $f$ tends to $+\infty$ when $x\rightarrow+\infty$.
A: Rearrange your inequality as $\frac {f'(x)} {1+f(x)^2} - \frac 1 {1+x^2} \geq 0$, which is equivalent to $(\arctan f(x) - \arctan x)' \geq 0$. Let $F(x)=\arctan f(x) - \arctan x$. Note that $F(0) = 0$ because $f(0)=0$. Your result will be an immediate consequence once we prove the following: "if $F' \geq 0$ and $F(0)=0$ then $F \geq 0$". But this is obvious since $F' \geq 0$ means that $F$ is increasing, so $F(x) \geq F(0) = 0 \space \forall x \geq 0$.
Therefore, $\arctan f(x) \geq \arctan x$ and since $\arctan$ is increasing, this means that $f(x) \geq x$, so $\lim \limits _{x \to \infty} f(x)= \infty$.
