## University Calculus: Early Transcendentals (3rd Edition)

a) $\sin 2 \theta$ and b) $\sin 2 \theta$
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 $m =\tan \theta$ $\dfrac{2 \tan \theta}{1+(\tan \theta)^2}=2 \sin \theta \cos \theta=\sin 2 \theta$