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- Thread starter asi123
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- #2

HallsofIvy

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[tex]\frac{x^2 y^2}{x^2y^2|(x-y)^2}[/tex]

I don't understand that "|".

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[tex]\frac{x^2 y^2}{x^2y^2|(x-y)^2}[/tex]

I don't understand that "|".

Where do u see a vertical line? it's a plus sign

I know that I need to show that [itex]\lim_{(x,y)\rightarrow (0,0)} f(x,y)= f(0,0)[/itex], any idea about how to show it?

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morphism

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Have you tried simplifying the expression for f(x,y) when (x,y)[itex]\neq[/itex](0,0)?I know that I need to show that [itex]\lim_{(x,y)\rightarrow (0,0)} f(x,y)= f(0,0)[/itex], any idea about how to show it?

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Have you tried simplifying the expression for f(x,y) when (x,y)[itex]\neq[/itex](0,0)?

What do you mean by simplify?

- #6

HallsofIvy

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[tex]\frac{x^2y^2}{x^2y^2+ (x- y)^2}[/tex]

is 1, the value of the function at (0,0).

There doesn't seem to me to be any good way to simplify that (multiplying out, reducing the fraction, etc. just like you did in algebra)

The best way to find the limit of a function of two variables, going to (0,0) is to change to polar coordinates. That way a single variable, r, measures the distance from (0,0). In polar coordinates, [itex]x= rcos(\theta)[/itex] and y= [itex]rsin(\theta)[/itex] so [itex]x^2y^2= r^4 sin^2(\theta)cos^2(\theta)[/itex] and [itex](x- y)^2= (rcos(\theta)- rsin(\theta))^2= r^2(cos(\theta)- sin(\theta))^2[/itex]. The function, in polar coordinates, is

[tex]\frac{r^4 sin^2(\theta)cos^2(\theta)}{r^2(sin^2(r^2theta)cos^2(\theta)sin^2(\theta)+ (cos(\theta)-sin(\theta))^2}= r^2\left[\frac{sin^2(\theta)cos^2(\theta)}{r^2sin^2(\theta)cos^2(\theta)+ (cos(\theta)- sin(\theta)^2}\right][/tex]

and now the point is that because of that "r

Now, that does cause a serious problem as far as showing that this function is continuous is concerned!

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[tex]\frac{x^2y^2}{x^2y^2+ (x- y)^2}[/tex]

is 1, the value of the function at (0,0).

There doesn't seem to me to be any good way to simplify that (multiplying out, reducing the fraction, etc. just like you did in algebra)

The best way to find the limit of a function of two variables, going to (0,0) is to change to polar coordinates. That way a single variable, r, measures the distance from (0,0). In polar coordinates, [itex]x= rcos(\theta)[/itex] and y= [itex]rsin(\theta)[/itex] so [itex]x^2y^2= r^4 sin^2(\theta)cos^2(\theta)[/itex] and [itex](x- y)^2= (rcos(\theta)- rsin(\theta))^2= r^2(cos(\theta)- sin(\theta))^2[/itex]. The function, in polar coordinates, is

[tex]\frac{r^4 sin^2(\theta)cos^2(\theta)}{r^2(sin^2(r^2theta)cos^2(\theta)sin^2(\theta)+ (cos(\theta)-sin(\theta))^2}= r^2\left[\frac{sin^2(\theta)cos^2(\theta)}{r^2sin^2(\theta)cos^2(\theta)+ (cos(\theta)- sin(\theta)^2}\right][/tex]

and now the point is that because of that "r^{2}" factored out, as r goes to 0, the whole thing goes to 0no matter what [itex]\theta[/itex] is.

Now, that does cause a serious problem as far as showing that this function is continuous is concerned!

Got ya, 10x.

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