Partial derivative limit

  • Thread starter ƒ(x)
  • Start date
  • #1
328
0

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
 

Answers and Replies

  • #2
127
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)
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
7
When you take the limit of a multivariable function, you have to do it along some path [itex]y(x)[/itex]...If every path leads to the same result, then the limit exists and is equal to that result.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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.
 
Last edited by a moderator:
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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
Science Advisor
Homework Helper
41,847
966
When you take the limit of a multivariable function, you have to do it along some path [itex]y(x)[/itex]...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 [itex]\theta[/itex], 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.)
 
  • #7
328
0
So does the limit for fy exist?
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,141
1,736
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}[/tex]
 
  • #9
328
0
The limit doesn't but the partial at that point does? Isn't it the otherway around?
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,141
1,736
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.
 
  • #11
328
0
So I was taking the wrong limit?
 
  • #12
HallsofIvy
Science Advisor
Homework Helper
41,847
966
With [itex]f(x,y)= (x^3+ y^3)^{1/3}[/itex] then [tex]f_y(0, 0)= \lim_{h\to 0}\frac{f(0, h)- f(0, 0)}{h}[/tex]. That is the limit you want to evaluate.

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

Related Threads on Partial derivative limit

  • Last Post
Replies
5
Views
769
Replies
3
Views
6K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
10
Views
800
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
1K
Top