# Partial derivative limit

## Homework Statement

f(x,y) = (x3+y3)^(1/3)

Show that fy(0,0) = 1

## The Attempt at a Solution

fy=y2/(x3+y3)^(2/3)

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

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
Homework Helper
Gold Member
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
Homework Helper
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.

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HallsofIvy
Homework Helper
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
$$\lim_{x\to 0}\frac{0^2}{(x^3)^{2/3}}= 0[/itex]. Since those limits are different, the limit of the function, as (x, y) goes to (0, 0), does not exist. HallsofIvy Science Advisor Homework Helper When you take the limit of a multivariable function, you have to do it along some path $y(x)$...If every path leads to the same result, then the limit exists and is equal to that result. The first part of that isn't true- you don't "have to do it along some path". The best way to evaluate limits in two dimensions, as (x, y) goes to (0, 0) is to change to polar coordinates. That way, r alone measures the distance to (0, 0). If the limit, as r goes to 0, does not depend on $\theta$, then the limit exists and is equal to that value. (If the limit point is not (0, 0) but, say, (a, b), translate your coordinate system by adding a to x and y to b.) So does the limit for fy exist? vela Staff Emeritus Science Advisor Homework Helper Education Advisor No, that limit doesn't exist, but the partial derivative does. Go back to the definition of the partial derivative: [tex]f_y(0,0) = \lim_{h \to 0} \frac{f(0,0+h)-f(0,0)}{h}$$

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

vela
Staff Emeritus
Homework Helper
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.