#### Answer

a) $\sin 2 \theta$ and b) $\sin 2 \theta$

#### Work Step by Step

a) We need to set $y=mx$ in the function.
$f(x,y)=\dfrac{2m}{1+m^2}$
Consider $ m =\tan \theta$
This implies that $f(x,y)=\dfrac{2m}{1+m^2}=\dfrac{2 \tan \theta}{1+(\tan \theta)^2}=2 \sin \theta \cos \theta=\sin 2 \theta$
b) Now, $\lim\limits_{(x,y) \to (0,0) } \dfrac{2 xy}{x^2+y^2}=\dfrac{2m}{1+m^2}$
From part (a), we set $ m =\tan \theta$
$\dfrac{2 \tan \theta}{1+(\tan \theta)^2}=2 \sin \theta \cos \theta=\sin 2 \theta$