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Question Number 187606 by otchereabdullai last updated on 19/Feb/23

$$Usepolarcoordinatetofind$$$$lim(x,y)\to (0,0)\frac{x^2-{xy}^2}{x^2+y^2}$$

Answered by a.lgnaoui last updated on 19/Feb/23

$$wecantrywithHopital$$$$rule$$$${lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}={lim}_{x\to a}\frac{f^{{}^{\prime }}\left(x\right)}{g^{{}^{\prime }}\left(x\right)}$$$$with:f\left(x\right)=x^2-xy$$$$g\left(x\right)=x^2+y^2$$$$$$

Answered by mr W last updated on 20/Feb/23

$$x=r\cos \theta$$$$y=r\sin \theta$$$$\frac{x^2-{xy}^2}{x^2+y^2}$$$$=\frac{r^2{\cos }^2\theta -r\cos \theta r^2{\sin }^2\theta }{r^2{\cos }^2\theta +r^2{\sin }^2\theta }$$$$={\cos }^2\theta -r\cos \theta {\sin }^2\theta$$$$\underset{x\to 0,y\to 0}{\lim }\frac{x^2-{xy}^2}{x^2+y^2}$$$$=\underset{r\to 0,\theta \to 0}{\lim }({\cos }^2\theta -r\cos \theta {\sin }^2\theta )$$$$={\cos }^{2}0-0$$$$=1$$

Commented by otchereabdullai last updated on 20/Feb/23

$$AmmuchgratefulprofW$$

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