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Question Number 192265 by yaslm last updated on 13/May/23

Answered by Frix last updated on 13/May/23

$$x=p+1$$$$y=q+1$$$$(p,q)\to (0,0)$$$$\frac{{(x-y)}^2}{x-y^2}=-\frac{{(p-q)}^2}{q^2-p+2q}$$$$p=r\cos \theta$$$$q=r\sin \theta$$$$r\to 0$$$$\frac{{(p-q)}^2}{q^2-p+2q}=\frac{(1-2\sin \theta \cos \theta )r}{r{\sin }^2\theta +2\sin \theta -\cos \theta }$$$$\Rightarrow$$$$\mathrm{Answer}\mathrm{is}0$$

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