Question

Given \(f(x,y) = \sqrt[3]{xy}\), is \(\frac{\partial f}{\partial x}\) continuous at (0,0)?

Answer 1

wouldn't that be a yes? trying the limit for x constant and y constant gives the same result

Answer 2

I mean the limit for y taking x =0 is 0, and the limit for x taking y=0 is 0, because the cube root of a negative number remains negative

Answer 3

But it is \(\frac{\partial f}{\partial x}\), not f(x,y)?

Answer 4

*facepalm*

Answer 5

no because x^(2/3) becomes 0 and the expression approaches inf? inky is uselesss

Answer 6

Hmm.. Aren't you getting 0/0 though?

Answer 7

that indeterminant form implies discontinuity right?

Answer 8

But someone insists that it is continuous...

Answer 9

find da book's answer key?

Answer 10

No

Answer 11

when does l'hopital's rule apply?

Answer 12

infty / infty, or 0/0. But I just know the single variable case. Not sure of the multi-variable case.

Answer 13

wolframy alpha?

Answer 14

Stumbles me as well,
\[\lim_{x\to0, y\to 0} (\frac{y}{x^2})^\frac{1}{3}\]

Answer 15

I'm here to learn, not to ask for an answer from WolframAlpha.

Answer 16

therefore, I'm going to go figure out the right answer from Wolfram and see if they show steps

Answer 17

Only if you can figure that out.

Answer 18

@skullpatrol Mind giving a hand here?

Answer 19

it's not continuous, but no steps.

Answer 20

If you want to learn about this, then this might be useful: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/limcont/limcont.html

Answer 21

I basically states that:
A function f of two variables is continuous at a point \((x_0,y_0)\) if:
\(f(x_0,y_0)\) is defined
\(\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)\) exists
\(\lim_{(x,y)\rightarrow (x_0,y_0)}f(x,y)=f(x_0,y_0)\)

Answer 22

*It basically...

Answer 23

Here's the problem.
I let g(x,y) = \(\frac{\partial f}{\partial x}\)
I thought when I needed to find the limit, it should be \(\lim_{(x,y)\rightarrow (0,0)}g(x,y)\)
However, someone told me that it should be \(\lim_{x\rightarrow 0}(\lim_{y\rightarrow 0}g(x,y))\)
I'm confused.

Answer 24

Look at the link I gave - it also has examples of how to calculate these limits

Answer 25

I know how to calculate the limits, I just don't understand what the right way to set up the limit is.

Answer 26

This video might help: http://www.youtube.com/watch?v=4-liutK41v0

Answer 27

Once again, the problem I have is to set up the limit, not to evaluate it.
When we find \(\frac{\partial f}{\partial x}\), we assume y as a constant. Then we need to find the limit based on our assumption that y is a constant. In this case, is it still appropriate to set up the limit like \(\lim_{(x,y) \rightarrow (0,0)} \frac{\partial f}{\partial x}\) ?

Answer 28

\[
\Large \frac{\partial f}{\partial x}=\frac{y}{3\sqrt[3]{x^2y^2}}=\frac{0}{0}\\

\Large \frac{\partial f}{\partial x}=\frac{y}{3\sqrt[3]{x^2y^2}}\frac{\sqrt[3]{x^4y^4}}{\sqrt[3]{x^4y^4}}=\frac{y\sqrt[3]{x^4y^4}}{3xy}=\frac{\sqrt[3]{x^4y^4}}{3x}=0
\]

Answer 29

\[\Large \frac{\partial f}{\partial x}=\frac{y}{3\sqrt[3]{x^2y^2}}=\frac{0}{0}\\

\Large \frac{\partial f}{\partial x}=\frac{y}{3\sqrt[3]{x^2y^2}}\frac{\sqrt[3]{x^4y^4}}{\sqrt[3]{x^4y^4}}=\frac{y\sqrt[3]{x^4y^4}}{3xy}=\frac{\sqrt[3]{x^4y^4}}{3x}=\frac{\sqrt[3]{x}\sqrt[3]{y^4}}{3}=0

\]

Answer 30

\[
\Large \frac{\partial f}{\partial x}=h(x,y)=\Large\frac{\sqrt[3]{x}\sqrt[3]{y^4}}{3}
\]

Answer 31

\[\Large \frac{\partial f}{\partial x}=\frac{y}{3\sqrt[3]{x^2y^2}}\frac{\sqrt[3]{x^4y^4}}{\sqrt[3]{x^4y^4}}=\frac{y\sqrt[3]{x^4y^4}}{3x^2y^2}=...?\]

Answer 32

@cinar I think something's wrong after the second equal sign in the second line of your post.

Answer 33

no, that's right..

Answer 34

\[\sqrt[3]{x^2y^2}\times \sqrt[3]{x^4y^4} = (x^2y^2)^\frac{1}{3}\times (x^4y^4)^\frac{1}{3} = [(x^2y^2)(x^4y^4)]^\frac{1}{3}=((xy)^6)^\frac{1}{3} = (xy)^2?\]

Answer 35

you are right
\[\Large \frac{\partial f}{\partial x}=h(x,y)=\Large\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}\\

\Large x=rcost\\
\Large y=rsint\\
\Large \lim_{(x,y)\rightarrow (0,0)}h(x,y)=\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}\\
\Large \lim_{r\rightarrow 0}\frac{\sqrt[3]{rsint}}{3\sqrt[3]{r^2cos^2t}}= \lim_{r\rightarrow 0}\frac{r^{\frac{1}{3}}\sqrt[3]{sint}}{3r^{\frac{2}{3}}\sqrt[3]{cos^2t}}=\lim_{r\rightarrow 0}\frac{r^{\frac{1}{3}}\sqrt[3]{sint}}{3r^{\frac{2}{3}}\sqrt[3]{cos^2t}}=\\
\Large \lim_{r\rightarrow 0}\frac{\sqrt[3]{sint}}{3r^{\frac{1}{3}}\sqrt[3]{cos^2t}}=0

\]
not continuous at (x,y)=(0,0)

Answer 36

\[
\Large \frac{\partial f}{\partial x}=h(x,y)=\Large\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}\\

\Large x=rcost\\
\Large y=rsint\\
\Large \lim_{(x,y)\rightarrow (0,0)}h(x,y)=\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}\\
\Large \lim_{r\rightarrow 0}\frac{\sqrt[3]{rsint}}{3\sqrt[3]{r^2cos^2t}}= \lim_{r\rightarrow 0}\frac{r^{\frac{1}{3}}\sqrt[3]{sint}}{3r^{\frac{2}{3}}\sqrt[3]{cos^2t}}=
\Large \lim_{r\rightarrow 0}\frac{\sqrt[3]{sint}}{3r^{\frac{1}{3}}\sqrt[3]{cos^2t}}=\infty
\]

Answer 37

@Callisto -- sorry I had to go eat something - am back now.

Remember that for a limit to exist, it must converge to the same value no matter path you take towards that limit (assuming the path is within the domain of the function).

Therefore, to calculate your limit:\[\lim_{(x,y)\rightarrow(0,0)}\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}\]you could chose a path that lies along \(y=\alpha x^2\) where \(\alpha\) is some constant. This gives:\[\lim_{(x,y)\rightarrow(0,0)}\frac{\sqrt[3]{y}}{3\sqrt[3]{x^2}}=\lim_{(x,y)\rightarrow(0,0)}\frac{\sqrt[3]{\alpha x^2}}{3\sqrt[3]{x^2}}=\lim_{(x,y)\rightarrow(0,0)}\frac{\sqrt[3]{\alpha}}{3}=g(\alpha)\]i.e. the limit depends on the value of \(\alpha\). Therefore the limit does not exist.

Answer 38

we are not approaching (0,0) from different directions. partial derivatives is a special case of directional derivative where y is constant or we only have a curve that is function of x.
I think we should consider one particular value of alpha than general value of it.

Answer 39

Could use the definition of limit itself to get:\[\frac{\partial f}{\partial x}(0,0)=\lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}\lim_{h\rightarrow0}\frac{0-0}{h}=\frac{0}{h}\]

Answer 40

*Could we use ...

Answer 41

yes that what i believe to be correct.

Answer 42

Hmmm... So this implies that \(\large\frac{\partial f}{\partial x}\) IS continuous at (0,0)?

Answer 43

i think it does ... there is a result on analysis that says that directional derivatives even if function is not continuous.

Answer 44

Interesting - looks like I need to do some reading :)

Answer 45

i didn't solve it. @Callisto did it herself.

Answer 46

I awarded the medal on the basis that you gave me some insight into something I was unaware of :)

Answer 47

@experimentX is just being humble. He guided me and explained the ideas to me.

Answer 48

the main point here is we all learnt something :)

Answer 49

lol ... still i learned something :)