$\displaystyle lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}$

when we put the x in the denominator inside the root

we get absolute value so wee need to split into cases

let us deal with only one case

$\displaystyle lim_{x->0^+}\frac{\sqrt{1+\frac{arctan}{x^2}}}{1}$

so i have a member in side the root which is 0/0

can i solve this thing without lhopital?