[tex]\displaystyle\lim_{(x,y)\to(1,2)}\dfrac{8x^2-2y^2}{2x-y}=2\lim_{(x,y)\to(1,2)}\dfrac{4x^2-y^2}{2x-y}=2\lim_{(x,y)\to(1,2)}\dfrac{(2x-y)(2x+y)}{2x-y}=\\\\=2\lim_{(x,y)\to(1,2)}2x+y=2\cdot(2\cdot1+2)=2\cdot4=8[/tex]
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[tex]\displaystyle\lim_{(x,y)\to(1,2)}\dfrac{8x^2-2y^2}{2x-y}=2\lim_{(x,y)\to(1,2)}\dfrac{4x^2-y^2}{2x-y}=2\lim_{(x,y)\to(1,2)}\dfrac{(2x-y)(2x+y)}{2x-y}=\\\\=2\lim_{(x,y)\to(1,2)}2x+y=2\cdot(2\cdot1+2)=2\cdot4=8[/tex]