## Calculus 8th Edition

$-1$
Conversion of polar co-ordinates $(r, \theta)$ are: $x=r \cos \theta$ and $y= r \sin \theta$ Given: $\lim\limits_{(x,y) \to(0,0)}\dfrac{e^{-x^2-y^2}-1}{(x^2+y^2)}$ This implies that $=\lim\limits_{r \to0}\dfrac{e^{(-r^2 \cos^2 \theta-r^2 \sin^2 \theta)}-1}{(r^2 \cos^2 \theta+r^2 \sin^2 \theta)}$ $=\lim\limits_{r \to0}\dfrac{e^{(-r^2 \cos^2 \theta-r^2 \sin^2 \theta)}-1}{r^2}$ Using L-Hospital's Rule. $:=\lim\limits_{r \to 0}\dfrac{-2re^{-r^2}}{2r}$ $=-1$