Question #249068

What is limit of (x,y) tends to (0,0) 1/sinxy?

Expert's answer

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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