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## Homework Statement

f(x,y) = (x

^{3}+y

^{3})^(1/3)

Show that f

_{y}(0,0) = 1

## The Attempt at a Solution

f

_{y}=y

^{2}/(x

^{3}+y

^{3})^(2/3)

And...I take the limit of it as x and y goes to zero, which gets me 0/0

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- #1

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f(x,y) = (x

Show that f

f

And...I take the limit of it as x and y goes to zero, which gets me 0/0

- #2

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eg, we can go along the path x=0, y=t. Sub that into f_y and calculate the limit as t->0.

Because it's a partial derivative, keeping x fixed, we would like to "fix x=0" first, and then calculus it in one dimension.

(not always, but it's something to try)

- #3

gabbagabbahey

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- #4

HallsofIvy

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The best way to take limits at (0, 0) for functions of two variables is to change into polar coordinates. That way, the single variable, r, measures the distance from (0, 0). If the limit, as r goes to 0, does **not** depend on [itex]\theta[/itex], then that is the limit as (x, y) goes to (0, 0).

Here, [tex]\frac{y^2}{(x^3+ y^3)^{2/3}}= \frac{r^2sin^2(\theta)}{r^3cos^3(\theta)+ r^3sin^3(\theta)}[/tex][tex]= \frac{r^2 sin^2(\theta)}{r^2(cos^3(\theta)+ sin^3(\theta))^{2/3}}[/tex]= [tex]\frac{sin^2(\theta)}{(cos^3(\theta)+ sin^3(\theta))^{2/3}}[/tex]

does not depend on r at all! The limit does not exist.

You could also have seen that by taking the limit as x goes to 0 first, then as y goes to 0: [tex]\lim_{y\to 0}\frac{y^2}{(y^3)^{2/3}}= \lim_{y\to 0}\frac{y^2}{y^2}= 1[/tex].

While taking the limit as y goes to 0 first, then as x goes to 0: [tex]\lim_{x\to 0}0= 0[/tex].

Since those limits are different, the limit as (x, y) goes to (0, 0) does not exist.

Here, [tex]\frac{y^2}{(x^3+ y^3)^{2/3}}= \frac{r^2sin^2(\theta)}{r^3cos^3(\theta)+ r^3sin^3(\theta)}[/tex][tex]= \frac{r^2 sin^2(\theta)}{r^2(cos^3(\theta)+ sin^3(\theta))^{2/3}}[/tex]= [tex]\frac{sin^2(\theta)}{(cos^3(\theta)+ sin^3(\theta))^{2/3}}[/tex]

does not depend on r at all! The limit does not exist.

You could also have seen that by taking the limit as x goes to 0 first, then as y goes to 0: [tex]\lim_{y\to 0}\frac{y^2}{(y^3)^{2/3}}= \lim_{y\to 0}\frac{y^2}{y^2}= 1[/tex].

While taking the limit as y goes to 0 first, then as x goes to 0: [tex]\lim_{x\to 0}0= 0[/tex].

Since those limits are different, the limit as (x, y) goes to (0, 0) does not exist.

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HallsofIvy

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If the limit exists (we don't yet know if it does, but the question posed seems to assume it does), then you can evaluate it by choosing a "path" to "lim" along.

No, that assumption is incorrect. Taking the limit as x goes to 0 first, then as y goes to 0 gives

[tex]\lim_{y\to 0}\frac{y^2}{(y^3)^{2/3}}= \lim_{y\to 0}\frac{y^2}{y^2}= 1[/tex]

but taking the limit as y goes to 0 first, then x goes to 0 gives

[tex]\lim_{x\to 0}\frac{0^2}{(x^3)^{2/3}}= 0[/itex].

Since those limits are different, the limit of the function, as (x, y) goes to (0, 0), does not exist.

- #6

HallsofIvy

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The first part of that isn't true- you

(If the limit point is not (0, 0) but, say, (a, b), translate your coordinate system by adding a to x and y to b.)

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So does the limit for f_{y} exist?

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vela

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[tex]f_y(0,0) = \lim_{h \to 0} \frac{f(0,0+h)-f(0,0)}{h}[/tex]

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The limit doesn't but the partial at that point does? Isn't it the otherway around?

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vela

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So I was taking the wrong limit?

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HallsofIvy

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(I see now that vela said that several posts before!)

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