Show that the function $y= f(x)= 1+ x^{2} + \arctan x^{2}$ is strictly monotone in $[0,+\infty ]$. If $f^{-1}$ is the reverse function of $f$, calculate the limit $$\lim_{y\rightarrow 1+}\frac{f^{-1}(y)}{\sqrt{y-1}}$$
I calculated the derivative of $f$ and I noticed that $f' > 0$ for every $x$ in $[0,+\infty ]$, so $f$ is strictly increasing.
But I have difficulties calculating the limit.