Range via Domain

For $x^2\ge 0,$ we can see that $\frac{x^2+1}{x^2+\sqrt{3}} = 1 + \frac{\sqrt{3}-1}{x^2+\sqrt{3}}$ is decreasing, ranging from $\frac{1}{\sqrt{3}}$ at $x=0$ and approaching 1 as $x\rightarrow\infty.$ Since $\tan^{-1}$ is an increasing function, with $\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$ and $\tan^{-1}(1) = \frac{\pi}{4},$ the range is $\left[\frac{\pi}{6},\frac{\pi}{4}\right).$

solve for $x=0~and~x\to \infty$ . Get the solution

this one is really interesting and easy . first find the minimum value of the expression and this will be when x is zero. and since a root three term is there in the denominator and 1 in the numerator. dr grows faster than the nr so it will always tend to approach the value of 1.