Answer
Limit does not exist.
Work Step by Step
The polar-coordinates are defined as: $x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$
$ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$
or, $=\lim\limits_{r \to 0} \dfrac{ 2 r \cos \theta}{r^2+r\cos \theta}$
or, $=\lim\limits_{r \to 0} \dfrac{ 2 \cos \theta}{r+\cos \theta}$
Suppose $ a \cos \theta \to 0$; $\lim\limits_{a \cos \theta \to 0} \dfrac{ 2 \cos \theta}{ cos \theta(a+1) }=\dfrac{2}{a+1}$
Then we can see that for this case, the limit is not a unique value.
Thus, $ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$ =Limit does not exist.
