# Help me evaluate this limit

1. Mar 23, 2008

### HF08

[SOLVED] Help me evaluate this limit

Part a:

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

Part b:

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

Lets work with Part a, shall we?

I seemed to be getting a $$\frac{0}{0}$$ 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) = $$\frac{(x^{2})^{2}+(y^{2})^{2}}{x^{2}+y^{2}}$$

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

2. Mar 23, 2008

### ircdan

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. Mar 24, 2008

### HallsofIvy

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

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

4. Mar 24, 2008

### HF08

Thanks

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. Mar 24, 2008

### HallsofIvy

Staff Emeritus
Yes, the limit for both is 0.