Mới thi hôm qua.

## December 28, 2007

## December 19, 2007

### Quen quen :-?

Let and . Prove that .

**Proof.** Let . For any , there is such that for , . Consider the set . If is empty, then the result immediately follows. So assume is nonempty and choose any . If , then , so which is a contradiction. Thus and . Then

or . Since is continuous, this inequality implies does not vanish at any point in .

Now assume there is a point and such that , Applying intermediate value theorem, we may assume is in the neighborhood of in . By mean value theorem, there is such that . Since cannot change its sign on , is strictly decreasing on . Especially, .

Now let , where on . Then

,

or

.

Since and , we have and is strictly decreasing. So must converge and we have

which contradicts that converges.

So, for all and , . This means is monotone increasing on . If for all , then must converge to some where . But this implies , leading to a contradiction. Hence for some sufficient large , and by our preceeding argument, for all . So as .

## December 13, 2007

### BĐT liên quan đến đạo hàm và quy tắc l’Hospital

Giả sử sao cho

với mọi . Chứng minh rằng .

*Lời giải*. Đặt

.

Khi đó ta có

.

Vì vậy

.

Tiếp tục đặt

.

Khi đó

.

Và vì vậy

.

Điều phải chứng minh.

## December 6, 2007

### Surface integral: The symmetric property

Why is for being the unit sphere.

,

by rotational symmetry of the sphere. The integral is

.

And the easiest way to find (should you want to) is again by symmetry

,

hence .