## Discontinuity

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A function $$f(x,y)$$ is continuous at a point $$(x_0,y_0)$$ if $\lim\limits_{(x,y)\to(x_0,y_0)} f(x,y) = f(x_0,y_0).$ For example, the function $f(x,y) = \dfrac{2xy}{x^2+y^2},$ is discontinuous at $$(0,0)$$ because $$f(0,0)$$ is not defined. If we look more closely, however, we will notice that the function does not have a limit at $$(0,0)$$ either, but we could indeed find a limit no matter with direction we choose to approach to $$(0,0)$$.

Suppose we approach $$(0,0)$$ along the direction which makes an angle $$\theta$$ with the $$x$$-axis. In this case the trajectory can be described by $$(x,y) = (t\cos\theta, t\sin\theta)$$ when $$t \to 0$$. Plugging it in $$f$$ gives $f(t\cos\theta, t\sin\theta) = \dfrac{2t^2\cos\theta \sin\theta}{t^2\cos^2\theta + t^2\sin^2\theta} = \sin 2\theta.$ As $$t\to 0$$, $$f \to \sin 2\theta$$ which depends on $$\theta$$. Since $$f$$ approaches to different values along different trajectories, we say that the limit $$\lim\limits_{(x,y)\to(0,0)} f(x,y)$$ does not exist.

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