# Finding the limit for this function of two variables

1. Oct 6, 2012

### lo2

1. The problem statement, all variables and given/known data

Ok I have this function:

$f(x,y)=\sqrt{4xy-3y^2}$

And then I have to find the limit:

$\lim_{h\rightarrow +0}{\frac{f(h,rh)}{h}}$ Where $r \in [0;\frac43]$

2. Relevant equations

3. The attempt at a solution

Well I must admit that I do not really understand the notation of this limit, does it mean that I have to replace x with h and y with rh, and then look at the limit? Or is it some kind of cylindric coordinate system, and if so where is the $\Theta$?

2. Oct 6, 2012

### tiny-tim

hi lo2!
yes, that's exactly what f(h,rh) means

3. Oct 6, 2012

### lo2

Hey thanks for the help!

I have come up with this answer:

$\lim_{h\rightarrow +0}{\frac{f(h,rh)}{h}}=$ Where $r \in [0;\frac43]$

$\frac{\sqrt{4h^2r-3r^2h^2}}{h}=\frac{h\sqrt{4r-3r^2}}{h}=\sqrt{4r-3r^2}= \sqrt{r(4-3r)}$

So this means that this limit does not depend on h really and that it only depends on the value of r. Would you deem this answer satisfactory?

4. Oct 6, 2012

### tiny-tim

yes, looks fine

(perhaps you should mention that h > 0, so it's always h, not -h, when you take h2 outside the square-root)

5. Oct 6, 2012

### SammyS

Staff Emeritus
To answer some of the questions in that last paragraph:

Expressing (x, y) as (h, rh), means that x = h, and y = rh. That's a parametric form of y = r x, so that's a line with a slope r that passes through the origin. Therefore, this has nothing to do with cylindrical (nor polar) coordinates.

In my opinion, that's a rather strange limit.

$\displaystyle \lim_{h\to 0^+}{\frac{f(h,rh)}{h}}$ is the limit of $\displaystyle \frac{f(x,y)}{x}$ as (x, y) → (0, 0) along a line with slope from 0 to 4/3 .

6. Oct 6, 2012

### lo2

Ok thanks! :) Yeah I probably should do that.

Btw if h approaches h from the left (h -> 0-) then the limit would be the negative of what I have found right?

7. Oct 6, 2012

right!

8. Oct 6, 2012

### lo2

Thanks Tiny Tim! I am glad you made it through a Christmas Carol ;)