Question #249068
What is limit of (x,y) tends to (0,0) 1/sinxy?
1
Expert's answer
2021-10-12T09:39:53-0400

Along x=yx=y


lim(x,y)(0,0)(1sin(xy))=lim(x,y)(0,0)(1sin(x2))=\lim\limits_{(x, y)\to(0,0)}(\dfrac{1}{\sin(xy)})=\lim\limits_{(x, y)\to(0,0)}(\dfrac{1}{\sin(x^2)})=\infin

Along x=yx=-y


lim(x,y)(0,0)(1sin(xy))=lim(x,y)(0,0)(1sin(x2))=\lim\limits_{(x, y)\to(0,0)}(\dfrac{1}{\sin(xy)})=\lim\limits_{(x, y)\to(0,0)}(\dfrac{1}{\sin(-x^2)})=-\infin

Therefore lim(x,y)(0,0)(1sin(xy))\lim\limits_{(x, y)\to(0,0)}(\dfrac{1}{\sin(xy)}) does not exist.


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