Finding the limit for this function of two variables

[lo2]

Homework Statement

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]

Homework Equations

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?

 

[tiny-tim]

hi lo2!

… 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?

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

 

[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}=<br /> \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?

 

[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)

 

[SammyS]

Homework Statement

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]

Homework Equations

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 cylindrical coordinate system, and if so where is the \theta?

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 .

 

[lo2]

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)

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?

 

[tiny-tim]

right!

 

[lo2]

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