# Bounding this expression

1. Nov 5, 2005

### twoflower

Hi,
could you please help me to show this expression goes to 0 az both $h_1$ and $h_2$ go to 0 and if I know that $\alpha > 3$?

$$\left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}\right|$$

Thank you!

2. Nov 5, 2005

### HallsofIvy

Staff Emeritus
First, remember that with two variables, it is not enough to just let one variable go to 0 and then the other. It is quite possible for a function to give two different limits if you approach (0,0) along two different paths. You need to show that if (x,y) is sufficiently close to (0,0), in any direction, then the function value will be 0.
I think it is best to transform the function to polar coordinates so that a single variable, r, measures the distance to the origin. That's especially easy in this problem: taking $h_1$ on the x-axis and $h_2$ on the y-axis, $h_1^2+ h_2^2= r^2$ and $h_1= r cos(\theta)$.
Put those into the formula and you will see immediately why that $\alpha> 3$ is necessary.

3. Nov 5, 2005

### benorin

I'll try

$$\lim_{h_1,h_2\rightarrow 0} \left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}}\right| = 0$$

if and only if

$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{h_{1}^{2}+h_{2}^{2}}<\delta \Rightarrow \left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}} - 0\right|<\epsilon$$

Since $\alpha > 3$, set $\alpha =3+\beta$ for $\beta >0$.

Choose $$\delta = \epsilon^{\frac{1}{\beta}}$$ so that if $$0<\sqrt{h_{1}^{2}+h_{2}^{2}}<\delta$$, then

$$\left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{h_1^2+h_2^2}} - 0\right|=\frac{\left|h_1\right|^{3+\beta}}{(h_1^2+h_2^2)^{\frac{3}{2}}}=\frac{\left|h_1\right|^{\beta} \left|h_1\right|^3 }{(h_1^2+h_2^2)^{\frac{3}{2}}}=\frac{\left|h_1\right|^{\beta} \left(h_1^2\right)^{\frac{3}{2}} }{(h_1^2+h_2^2)^{\frac{3}{2}}} \leq \frac{\left|h_1\right|^{\beta} \left(h_1^2+h_2^2\right)^{\frac{3}{2}} }{(h_1^2+h_2^2)^{\frac{3}{2}}} = \left|h_1\right|^{\beta} = \left(h_1^2\right)^{\frac{\beta}{2}} \leq \left(h_1^2+h_2^2\right)^{\frac{\beta}{2}}$$
$$= \left[\left(h_1^2+h_2^2\right)^{\frac{1}{2}} \right]^{\beta} < \delta^{\beta} = \left(\epsilon^{\frac{1}{\beta}}\right)^{\beta} = \epsilon$$

Last edited: Nov 5, 2005
4. Nov 5, 2005

### benorin

It might be better (in the above proof) to replace $\beta$ with $\alpha -3$.

5. Nov 5, 2005

### twoflower

Thank you HallsoftIvy, I tried it and is surprisingly easy to prove the limit is zero using polar expression of $h_1$ and $h_2$.

There's just one thing I'm not completely sure about yet:
why is it correct to put

$$h_1^2 + h_2^2 = r^2$$

? I see it is suitable to look at $h_1$ and $h_2$ as they lie on common circle with radius going zero (and thus both points going to 0), but aren't we ommiting something within two-variable functions problematics?

I don't know whether I expressed myself in an understandable way...

6. Nov 5, 2005

### HallsofIvy

Staff Emeritus
Having set up a Cartesian coordinate system with x= h1 and y= h2, then polar coordinates are DEFINED by
$$r= \sqrt{h_1^2+h_2^2$$ and $$\theta= arctan(\frac{h_2}{h_1})$$
Conversely,
$$h_1= r \cos(\theta)$$
Would the "something" you feel is being omitted be the angle $\theta$?

It is not omitted. Your formula
$$\left| \frac{\left|h_1\right|^\alpha}{(h_1^2+h_2^2)\sqrt{ h_1^2+h_2^2}}\right|$$
in polar coordinates is
$$\left|\frac{r^\alpha cos^\alpha(\theta)}{r^3}\right|$$
which is not only simpler but has the crucial property that if $\alpha> 3$, then

9. Nov 5, 2005

### twoflower

Yes, so the right answer for $$\alpha = 3$$ would be that the limit does exist, but depends on $$\theta$$, not that it doesn't exist as I understood it from HallsoftIvy's last response..

10. Nov 5, 2005

### benorin

But the limit as h1,h2->0 only implies that r->0, theta must be free to vary since h1 and h2 may approach the origin along any curve. If the value of theta affects the value of the limit in polar coordinates, then the limit does not exist is rectangular coordinates, as such would indicate that the limit has different values depending on how (along what curve) h1,h2->0.

11. Nov 5, 2005

### twoflower

Thank you benorin, that's what I was asking for, now I hope I'm sure about that matter, so anytime $\theta$ would affect value of limit, it doesn't exist.

This is just the same case as in rectangular coordinates - when I approach to origin with one variable directly and with the second one on straight line and the limit would depend on $k$ (if let's say $y = kx$).

12. Nov 5, 2005

### benorin

Exactly

Exactly.

Yes, in the case that k effects the value of the limit, then the limit d.n.e. (does not exist), for example: the case when the value of the limit when approaching along the negative x-axis and the value of the limit when approaching along the positive y-axis are not equal, the the limit d.n.e. (e.g., if $\lim_{x\rightarrow 0^{-}}f(x,0)=-3$, and if $\lim_{y\rightarrow 0^{+}}f(0,y)=0$, then $\lim_{(x,y)\rightarrow (0,0)}f(x,y)=d.n.e.$.)

This sort of thing would correspond to the case where the polar limit would vary with $\theta$, (and hence the limit d.n.e.); in the above example: the limit when approaching along the negative x-axis would be with $\theta = \pi$, and the value of the limit when approaching along the positive y-axis would be with $\theta = \frac{\pi}{2}$, (e.g., if $\lim_{r\rightarrow 0^{+}}f(r\cos(\pi),r\sin(\pi))=-3$, and if $\lim_{r\rightarrow 0^{+}}f(r\cos(\frac{\pi}{2}),r\sin(\frac{\pi}{2}))=0$, then $\lim_{r\rightarrow 0^{+}}f(r\cos(\theta),r\sin(\theta))=d.n.e.$.)

Note that: $\lim_{r\rightarrow 0^{+}}f(r\cos(\theta),r\sin(\theta))=\lim_{(x,y)\rightarrow (0,0)}f(x,y)$

Last edited: Nov 5, 2005
13. Nov 5, 2005

### HallsofIvy

Staff Emeritus
Yes, but not relevant. For some other $\theta$ value, the expression will be non-zero. Therefore, the limit itself does not exist.