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Question Number 191733 by Spillover last updated on 29/Apr/23

$$Showthat$$$$\underset{(x,y)\to (0,0)}{\lim }\frac{x^2-y^2}{x^2+y^2}doesnotexist$$

Answered by mehdee42 last updated on 30/Apr/23

$$lety=x\to {lim}_{(x,y)\to (0,0)}\frac{x^2-y^2}{x^2+y^2}=0$$$$lety=2x\to {lim}_{(x,y)\to (0,0)}\frac{x^2-y^2}{x^2+y^2}=-\frac{3}{5}$$$$thereforthislimitdoesnotexist$$$$$$$$$$

Answered by Frix last updated on 29/Apr/23

$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}r\cos \theta \\ r\sin \theta \end{array}\right)$$$$\underset{\left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}0\\ 0\end{array}\right)}{\lim }\frac{x^2-y^2}{x^2+y^2}=\underset{r\to 0}{\lim }\frac{r^2{\cos }^2\theta -r^2{\sin }^2\theta }{r^2{\cos }^2\theta +r^2{\sin }^2\theta }=$$$$=\cos 2\theta \Rightarrow \mathrm{the}\mathrm{limit}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{as}\mathrm{its}$$$$\mathrm{value}\mathrm{depends}\mathrm{on}\mathrm{the}\mathrm{direction}$$

Commented by Spillover last updated on 30/Apr/23

$$thankyou$$

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