## Thomas' Calculus 13th Edition

a) $\sin 2 \theta$ b) $\sin 2 \theta$
a) Substitute $y=mx$ in the function, so we have: $f(x,y)=\dfrac{2m}{1+m^2}$ Now, set $m =\tan \theta$ Thus, $f(x,y)=\dfrac{2m}{1+m^2}=\dfrac{2 \tan \theta}{1+(\tan \theta)^2}$ or, $f(x,y)=2 \sin \theta \cos \theta=\sin 2 \theta$ b) $\lim\limits_{(x,y) \to (0,0) } \dfrac{2 xy}{x^2+y^2}=\dfrac{2m}{1+m^2}$ Need to consider part (a), set $m =\tan \theta$ $\dfrac{2 \tan \theta}{1+(\tan \theta)^2}=\sin 2 \theta$