Show-that-lim-x-y-0-0-x-2-y-2-x-2-y-2-does-not-exist-

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

S h o w t h a t

\lim_{(x, y) \to (0, 0)} \frac{x^{2} - y^{2}}{x^{2} + y^{2}} d o e s n o t e x i s t

Answered by mehdee42 last updated on 30/Apr/23

l e t y = x \to {l i m}_{(x, y) \to (0, 0)} \frac{x^{2} - y^{2}}{x^{2} + y^{2}} = 0

l e t y = 2 x \to {l i m}_{(x, y) \to (0, 0)} \frac{x^{2} - y^{2}}{x^{2} + y^{2}} = - \frac{3}{5}

t h e r e f o r t h i s l i m i t d o e s n o t e x i s t

Answered by Frix last updated on 29/Apr/23

(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} r \cos θ \\ r \sin θ \end{matrix})

\lim_{(\begin{matrix} x \\ y \end{matrix}) \to (\begin{matrix} 0 \\ 0 \end{matrix})} \frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \lim_{r \to 0} \frac{r^{2} \cos^{2} θ - r^{2} \sin^{2} θ}{r^{2} \cos^{2} θ + r^{2} \sin^{2} θ} =

= \cos 2 θ \Rightarrow the limit does not exist as its

value depends on the direction

Commented by Spillover last updated on 30/Apr/23

t h a n k y o u