Question

Question on comment
Please, help

Answer 1

hello loser what do you need help with?

Answer 2

omg this sounds so weird i don't actually mean loser ...

Answer 3

\(f(x,y) =\begin{cases}(x^2+y^2) sin (\dfrac{1}{x^2+y^2}~~~if~~~ (x,y) \neq (0,0)\\0~~~if~~~(x,y) =(0,0)\end{cases}\)

Answer 4

Prove: f(x,y) is differentiable at (0,0) but \(\dfrac{\partial f}{\partial x}\), \(\dfrac{\partial f}{\partial y}\) are not continuous at(0,0)

Answer 5

See http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives

Answer 6

@Loser66

Answer 7

Thank you. :)

Answer 8

I got all, but for the part
\(\dfrac{\partial f}{\partial x}(x,y)= lim_{(x,y)\rightarrow (0,0)}(2xsin(\dfrac{1}{x^2+y^2}-\dfrac{2x}{x^2+y^2}cos(\dfrac{1}{x^2+y^2}))\)
I need a neat argument to say it is undefined

Answer 9

@pitamar

Answer 10

turn to polar: \(x =r cos \theta\)
\(y = r sin \theta\)
as \((x,y)\rightarrow (0,0)\) , \(r\rightarrow 0\)
lt becomes

Answer 11

\(lim_{r\rightarrow 0} 2rcos \theta sin(1/r^2) -\dfrac{2 cos (1/r^2)}{r}\)

Answer 12

Question: \(lim_{r\rightarrow 0} \dfrac{1}{r}=??\) \(\infty\) or \(undefined\)??
I know I should not be confused on this basic knowledge like that but ..... I have no choice. hehehe

Answer 13

\[
f(x,y)=\left(x^2+y^2\right) \sin
\left(\frac{1}{x^2+y^2}\right)
\]

\[
g(x,y)=\frac{\partial f(x,y)}{\partial x}=2 x \sin \left(\frac{1}{x^2+y^2}\right)-\frac{2 x \cos
\left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}
\]


\[
g\left(\frac{1}{\sqrt{\pi n}},0\right)=\\\frac{2 \sin (\pi n)}{\sqrt{\pi } \sqrt{n}}-2 \sqrt{\pi
} \sqrt{n} \cos (\pi n)=-2 \sqrt{\pi } \sqrt{n} \cos (\pi n)=\\-2 \sqrt{\pi } \sqrt{n} (-1)^n\]

Answer 14

This will do it

Answer 15

@Loser66

Answer 16

I don't get why we have to do it and how it relates to our argument. I am sorry for being dummy.

Answer 17

You have a sequence
\[\left (\frac {1} {\sqrt {\pi n}}, 0 \right) -> (0, 0)\]

Answer 18

but
\[g\left(\frac{1}{\sqrt{\pi n}},0\right)=\\\frac{2 \sin (\pi n)}{\sqrt{\pi } \sqrt{n}}-2 \sqrt{\pi
} \sqrt{n} \cos (\pi n)=\\-2 \sqrt{\pi } \sqrt{n} \cos (\pi n)=-2 \sqrt{\pi } \sqrt{n} (-1)^n
\]
does not go to zero, so the g is not continuous at (0,0). Got it

Answer 19

Yes, thank you so much. :)