I have read I can replace in limits $\arctan x$ with $x$. I conjectured that I can replace $\arctan x^2$ with $x^2$, however in following example it doesn't work:
$$\frac{2x\left( 105x^4+150 x^2 \left(-1+3 x^4\right)\arctan x^2+45(\arctan x^2)^2\right)}{\left(1+x^4\right)^5 (\arctan x^2)^{9/2}}$$
$x$ going to zero expression is 940 but replacing $\arctan x^2$ with $x^2$ limit is 900.
Why result is incorrect? Maybe I should replace $\arctan x^2$ with other function?
Note that replacing $\arctan x$ with $x$ doesn't give you an exact answer - basically, you are Taylor expanding $\arctan x$ about $x=0$: $$\arctan x\approx x-\dfrac{x^3}{3}+\dfrac{x^5}{5}\\ \implies \arctan x^2\approx x^2-\dfrac{x^6}{3}+\dfrac{x^{10}}{5}$$ Hence, the answer you get by replacing $\arctan x$ with $x$ (and replacing $\arctan x^2$ with $x^2$) is an approximation to the actual limit.