Limit of fy as x and y approach zero

[ƒ(x)]

Homework Statement

f(x,y) = (x³+y³)^(1/3)

Show that f_y(0,0) = 1

The Attempt at a Solution

f_y=y²/(x³+y³)^(2/3)

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

 

[Jerbearrrrrr]

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.
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)

 

[gabbagabbahey]

When you take the limit of a multivariable function, you have to do it along some path $y(x)$...If every path leads to the same result, then the limit exists and is equal to that result.

 

[HallsofIvy]

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 $\theta$, then that is the limit as (x, y) goes to (0, 0).

Here, $$\frac{y^2}{(x^3+ y^3)^{2/3}}= \frac{r^2sin^2(\theta)}{r^3cos^3(\theta)+ r^3sin^3(\theta)}$$$$= \frac{r^2 sin^2(\theta)}{r^2(cos^3(\theta)+ sin^3(\theta))^{2/3}}$$= $$\frac{sin^2(\theta)}{(cos^3(\theta)+ sin^3(\theta))^{2/3}}$$
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: $$\lim_{y\to 0}\frac{y^2}{(y^3)^{2/3}}= \lim_{y\to 0}\frac{y^2}{y^2}= 1$$.

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

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

 

[HallsofIvy]

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

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].<br /> <br /> Since those limits are different, the limit of the function, as (x, y) goes to (0, 0), does not exist.[/tex]

 

[HallsofIvy]

When you take the limit of a multivariable function, you have to do it along some path $y(x)$...If every path leads to the same result, then the limit exists and is equal to that result.

The first part of that isn't true- you don't "have to do it along some path". The best way to evaluate limits in two dimensions, as (x, y) goes to (0, 0) is to change to polar coordinates. That way, r alone measures the distance to (0, 0). If the limit, as r goes to 0, does not depend on $\theta$, then the limit exists and is equal to that value.

(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.)

 

[ƒ(x)]

So does the limit for f_y exist?

 

[vela]

No, that limit doesn't exist, but the partial derivative does. Go back to the definition of the partial derivative:

$$f_y(0,0) = \lim_{h \to 0} \frac{f(0,0+h)-f(0,0)}{h}$$

 

[ƒ(x)]

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

 

[vela]

If the limit in post 8 doesn't exist, then the partial derivative doesn't exist. The limit you tried to calculate in the original post, however, is a different limit. Whether or not it exists says nothing about whether the partial derivative exists.

 

[ƒ(x)]

So I was taking the wrong limit?

 

[HallsofIvy]

With $f(x,y)= (x^3+ y^3)^{1/3}$ then $$f_y(0, 0)= \lim_{h\to 0}\frac{f(0, h)- f(0, 0)}{h}$$. That is the limit you want to evaluate.

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