Help me evaluate this limit

  • Thread starter HF08
  • Start date
  • #1
39
0
[SOLVED] Help me evaluate this limit

Part a:

Show that f(x,y) = [tex]\frac{x^{4}+y^{4}}{x^{2}+y^{2}}[/tex]
as (x,y) -> (0,0)


Part b:

Similarily, show that f(x,y) = [tex]\frac{\sqrt{\left|xy\right|}}{\sqrt[3]{x^{2}+y^{2}}}[/tex] as (x,y)->(0,0)

Lets work with Part a, shall we?

I seemed to be getting a [tex]\frac{0}{0}[/tex] type limit. In prior situations and examples, I have seen clever substitutions or inequality relationships to aid us. In this spirit of such an approach I have thought it obvious to write it as:

f(x,y) = [tex]\frac{(x^{2})^{2}+(y^{2})^{2}}{x^{2}+y^{2}}[/tex]

Unfortunate for me is that this rather unoriginal form lends me nothing (at least to me). I have the sum of squares, not the difference of squares. I have thought of multiplying both the numerator and denominator by an indentity of some kind, but this might also be provide poor results? Can you please help me here?

As for part b, I am sure similar strategies are at work. I am also aware that part b looks acutely more difficult. Hence, please help me work on part a first. It isn't lack of effort, I assure you as my post here implies.

Thank you,
HF08
 

Answers and Replies

  • #2
229
0
I will help you with the first one to give you an idea of how to do these.

So let e > 0. Choose d = (fill this in). Then if 0 <|(x,y) - (0, 0)| < d (ie, sqrt(x^2 + y^2) < d), we have

|(x^4 + y^4)/(x^2 + y^2)| <= x^4/|x^2 + y^2| + y^4/|x^2 + y^2| <= x^2 + y^2 < d^2


Just think about why every step is true and pick d appropriately to make things work.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
When you have two variables, often the simplest way to do a limit is to change to polar coordinates: with [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex],
your first fraction becomes
[tex]\frac{r^4cos^4(\theta)+ r^4sin^4(\theta)}{r^2}= r^2(cos^4(\theta)+ sin^4(\theta))[/tex]
Now, (x,y) going to (0,0) mean r goes to 0 with [itex]\theta[/itex] indeterminate. Can you see that the limit, as r goes to 0, does not depend on [itex]\theta[/itex]?

For the second, again change to polar coordinates:
[tex]\frac{\sqrt{r^2(cos\theta sin\theta)}}{r^{\frac{2}{3}}}[/tex]
Again, the limit, as r goes to 0, does not depend on [itex]\theta[/itex].
 
  • #4
39
0
Thanks

When you have two variables, often the simplest way to do a limit is to change to polar coordinates: with [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex],
your first fraction becomes
[tex]\frac{r^4cos^4(\theta)+ r^4sin^4(\theta)}{r^2}= r^2(cos^4(\theta)+ sin^4(\theta))[/tex]
Now, (x,y) going to (0,0) mean r goes to 0 with [itex]\theta[/itex] indeterminate. Can you see that the limit, as r goes to 0, does not depend on [itex]\theta[/itex]?

For the second, again change to polar coordinates:
[tex]\frac{\sqrt{r^2(cos\theta sin\theta)}}{r^{\frac{2}{3}}}[/tex]
Again, the limit, as r goes to 0, does not depend on [itex]\theta[/itex].


Excellent post HallsofIvy. I have found both limits to be 0. This is more useful than the definition approach. If you agree with what the limit is, I will mark this solved.

Thank You,
HF08
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Yes, the limit for both is 0.
 

Related Threads on Help me evaluate this limit

  • Last Post
Replies
6
Views
2K
Replies
7
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
1
Views
947
  • Last Post
Replies
6
Views
1K
Replies
9
Views
1K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
17
Views
8K
  • Last Post
Replies
9
Views
2K
Top