Multivariable limit definition question

1. Oct 15, 2012

Zondrina

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

I'm reading through Taylor's advanced calculus and came across this question in section 7.2 :

http://gyazo.com/6b0c5a2e4e605ff77bf6584eb3295948

2. Relevant equations

The definition of the partial of f with respect to some variable at some point (a,b), lets say the partial with respect to x of some f(x,y) would be :

$\frac{∂f}{∂x} (a, b) = \lim_{h→0} \frac{f(a+h, b) - f(a, b)}{h}$

3. The attempt at a solution

Okay, so the question wants me to compute the second partials given the conditions on f.

So first I note that f is continuous everywhere for x,y ≠ 0. I also note that f(x,0) = f(0,y) = f(0,0) = 0 has been defined nicely for me as to make my function continuous at (0,0).

I suppose first that I'll show that $\frac{∂^2f}{∂y∂x} = -1$. So I first have to do a subsidiary calculation to find the first partial with respect to x.

$\frac{∂f}{∂x} (0, 0) = \lim_{h→0} \frac{f(h, 0) - f(0, 0)}{h} = \lim_{h→0} \frac{0 - 0}{h} = 0$

Okay, that was not so bad since f was nicely defined for us. Now here comes my problem when I try to find the second partial. Note that when I use subscripts I refer to the partial of the function with respect to that variable. So $f_x = \frac{∂f}{∂x}$.

$\frac{∂^2f}{∂y∂x} (0, 0) = \lim_{k→0} \frac{f_x(0, k) - f_x(0, k)}{k} (***)$

I'm labeling this equation (***) so I can return to it after I calculate fx(0,k). Now to calculate it :

$\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - h^2arctan(h/k)}{h} = 0$

Now this produces a big problem when I plug it back into (***) as I don't get the answer of -1 which I'm apparently supposed to get :

$\frac{∂^2f}{∂y∂x} (0, 0) = \lim_{k→0} \frac{f_x(0, k) - f_x(0, k)}{k} = \lim_{k→0} \frac{0 - 0}{k} = 0$

What am I doing wrong here? Have I missed something or have I done a calculation wrong?

2. Oct 15, 2012

micromass

Staff Emeritus
I think there is a typo here. The last limit should be

$$\lim_{h\rightarrow 0} \frac{h^2arctan(k/h)-k^2arctan(h/k)}{h}$$

This limit is not zero, I think.

3. Oct 15, 2012

Zondrina

You sir, have made my day. That will teach me to speed through my calculations too quickly haha.

So really I would get :

$\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - k^2arctan(h/k)}{h}$

and switching to polar co-ordinates ill presume will give me my desired result?

4. Oct 15, 2012

micromass

Staff Emeritus
Yes, try to work in polar coordinates. I think that'll make things easy.

5. Oct 15, 2012

Zondrina

Hmm I've hit a hitch in my calculations again :

For h=rcosθ and k = rsinθ we get :

$\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - k^2arctan(h/k)}{h} = \lim_{r→0} \frac{r^2cos^2θarctan(tanθ) - r^2sin^2arctan(cotθ)}{rcosθ}$

Which tends to 0 as r tends to 0?

6. Oct 15, 2012

micromass

Staff Emeritus
Hmmm, maybe going over in polar coordinates was not a good idea after all. The problem is that $h\rightarrow 0$ doesn't exactly correspond to $r\rightarrow 0$.

I actually don't think it is too hard to calculate the original limit explicitely:

$$\lim_{h\rightarrow 0} \frac{h^2arctan(k/h)-k^2 arctan(h/k)}{h^2}=\lim_{h\rightarrow 0} h \cdot arctan(k/h) - k^2\lim_{h\rightarrow 0} \frac{arctan(h/k)}{h}$$

The first limit is 0 (why?). The second limit can be calculated by l'Hopitals rule.

7. Oct 15, 2012

Zondrina

The first limit is 0 because of the squeeze theorem. After applying L'Hospital's rule I get (2h)arctan(k/h) - k and as h→0, (2h)arctan(k/h) - k → -k just like I wanted.