- #1
adoion
- 55
- 0
Hi,
[tex]f(X)=\frac{xy^2}{x^2+y^4}[/tex] is the function in question, this is the value of the function at ##X=(x,y)## when ##x\neq0##, and ##f(X)=0## when ##X=(0,y)## for any ##y## even ##y=0##.
Now, along any vector or line from the origin the directional derivative ##f'(Y,0)## (where ##Y=(a,b)## is the vector or line along witch the derivative is being taken and ##0## is the point at witch it is taken, in this case.) exists for any vector ##Y=(a,b)## (this is easily verifiable).
But the function is not continuous at ##X=(0,0)##, since the value of the function is ##0## at ##X=(0,0)## but along the parabola ##x=y^2## the value of the function is ##\frac{1}{2}## and the parabola comes arbitrary close to the value ##(0,0)##.
So how is this possible I am trying to get my head around this but can't figure it out??
[tex]f(X)=\frac{xy^2}{x^2+y^4}[/tex] is the function in question, this is the value of the function at ##X=(x,y)## when ##x\neq0##, and ##f(X)=0## when ##X=(0,y)## for any ##y## even ##y=0##.
Now, along any vector or line from the origin the directional derivative ##f'(Y,0)## (where ##Y=(a,b)## is the vector or line along witch the derivative is being taken and ##0## is the point at witch it is taken, in this case.) exists for any vector ##Y=(a,b)## (this is easily verifiable).
But the function is not continuous at ##X=(0,0)##, since the value of the function is ##0## at ##X=(0,0)## but along the parabola ##x=y^2## the value of the function is ##\frac{1}{2}## and the parabola comes arbitrary close to the value ##(0,0)##.
So how is this possible I am trying to get my head around this but can't figure it out??