calculating the limit of a 2 variable function

Question

The function is $$f(x,y) = \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$$ Prove (and calculate) / refute the Limit of the function where $x$ and $y$ go to $0 (x\rightarrow 0$ and $y\rightarrow 0)$

Answer 1

Hint: In polar coordinates we have

$$f(r\cos t, r\sin t) =\frac{e^{-1/r^2}}{r^4(\cos^4t + \sin^4 t)}.$$

Argue that the denominator above is $\ge r^4 c$ for some positive $c.$

Answer 2

Let polar coordinates $\\ x=r\cos { \theta } \\ y=r\sin { \theta } $ then we have $$\lim _{ x\rightarrow 0,y\rightarrow 0 }{ \frac { { e }^{ -\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } } }{ { x }^{ 4 }+{ y }^{ 4 } } } =\lim _{ r\rightarrow 0 }{ \frac { { e }^{ -\frac { 1 }{ { r }^{ 2 } } } }{ { r }^{ 4 }\left( \cos ^{ 4 }{ \theta } +\sin ^{ 4 }{ \theta } \right) } } =\frac { 1 }{ \left( \cos ^{ 4 }{ \theta } +\sin ^{ 4 }{ \theta } \right) } \lim _{ r\rightarrow 0 }{ \frac { { e }^{ -\frac { 1 }{ { r }^{ 2 } } } }{ { r }^{ 4 } } } \\ \\ \\ $$

the last limit is equal to $0$ , you can observe it by the graphic of the function of $\frac { { e }^{ -\frac { 1 }{ { r }^{ 2 } } } }{ { r }^{ 4 } } $