[tex]\displaystyle\\\lim_{(x,y)\to(0,0)}\dfrac{(\sin x+ \cos y)^2-\cos ^2 y}{\sin x \cos y}=\lim_{(x,y)\to(0,0)}\dfrac{\sin^2 x+2\sin x \cos y+ \cos^2 y-\cos ^2 y}{\sin x \cos y}=\\\\=\lim_{(x,y)\to(0,0)}\dfrac{\sin^2 x+2\sin x \cos y}{\sin x \cos y}=\lim_{(x,y)\to(0,0)}\dfrac{\sin x+2\cos y}{\cos y}=\dfrac{\sin 0 +2\cos 0}{\cos 0}=\\=\dfrac{0+2\cdot1}{1}=2[/tex]
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[tex]\displaystyle\\\lim_{(x,y)\to(0,0)}\dfrac{(\sin x+ \cos y)^2-\cos ^2 y}{\sin x \cos y}=\lim_{(x,y)\to(0,0)}\dfrac{\sin^2 x+2\sin x \cos y+ \cos^2 y-\cos ^2 y}{\sin x \cos y}=\\\\=\lim_{(x,y)\to(0,0)}\dfrac{\sin^2 x+2\sin x \cos y}{\sin x \cos y}=\lim_{(x,y)\to(0,0)}\dfrac{\sin x+2\cos y}{\cos y}=\dfrac{\sin 0 +2\cos 0}{\cos 0}=\\=\dfrac{0+2\cdot1}{1}=2[/tex]