- #1
eptheta
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I have a function in x and y, and I was trying to figure out if it was continuous or not.
[itex]f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}[/itex]
As far as I know, the only problematic point in the domain is (x,y)=(0,0) so I tried to use the [itex] \epsilon,\delta [/itex] definition.
My proposed limit at (0,0) being 0,
[itex]\frac{|x^2-y^2|}{(x^2+y^2)^2} < \epsilon [/itex]
when [itex]\sqrt{x^2+y^2}<\delta[/itex]
The epsilon term [itex]< \frac{|x^2|+|y^2|}{(x^2+y^2)^2}
=\frac{x^2+y^2}{(x^2+y^2)^2}
=\frac{1}{(x^2+y^2)} = \frac{1}{\delta^2}[/itex]
therefore, take [itex]\delta=\frac{1}{\sqrt{\epsilon}}
[/itex] and the definition is fulfilled.
1.Wolfram alpha tells me the limit doesn't exist, the plot clearly shows me the graph tends towards infinity at (0,0), so there's something wrong with my concepts.
Could someone point out what ?
2.Is the fact that the limit exists at every point a sufficient condition for continuity?
Thanks.
[itex]f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}[/itex]
As far as I know, the only problematic point in the domain is (x,y)=(0,0) so I tried to use the [itex] \epsilon,\delta [/itex] definition.
My proposed limit at (0,0) being 0,
[itex]\frac{|x^2-y^2|}{(x^2+y^2)^2} < \epsilon [/itex]
when [itex]\sqrt{x^2+y^2}<\delta[/itex]
The epsilon term [itex]< \frac{|x^2|+|y^2|}{(x^2+y^2)^2}
=\frac{x^2+y^2}{(x^2+y^2)^2}
=\frac{1}{(x^2+y^2)} = \frac{1}{\delta^2}[/itex]
therefore, take [itex]\delta=\frac{1}{\sqrt{\epsilon}}
[/itex] and the definition is fulfilled.
1.Wolfram alpha tells me the limit doesn't exist, the plot clearly shows me the graph tends towards infinity at (0,0), so there's something wrong with my concepts.
Could someone point out what ?
2.Is the fact that the limit exists at every point a sufficient condition for continuity?
Thanks.