Intersecting Curves

The Range of function $x^2 \geq 0$ and it continius function.

The value of $\lim_{x \rightarrow \pm \infty } \frac{1}{1+x^2} =1,$ it's means the asymtot of function $y= \frac{1}{1+x^2}$ is $y=1$ and have intersection with-y at point $(0,1).$

So, both functions will intersection if: $y= \frac{1}{1+y} \implies y^2+y=1 \implies y^2+y-1=0 \implies y= \frac{-1 + \sqrt5}{2}$ (Because $y= \frac{-1 - \sqrt5}{2} < 0$ ).

Two points of intersection both functions are $( \pm { \sqrt {\frac{-1 +\sqrt5}{2}}}, \frac{-1 + \sqrt5}{2} )$ .