Is unbounded function multiplied by bounded function or unbounded function?

Is unbounded function multiplied by bounded function or unbounded function?

It can be bounded or unbounded
[strictly declare that all function fields discussed below are non negative]
For example, the function y = x is unbounded, the function y = 1 / X is bounded, and their deposition is y = 1 bounded
For example, the function y = x * x is unbounded, the function y = 1 / X is bounded, and their deposition is y = x unbounded
Generally speaking, when an unbounded function tends to ∞, its value also tends to ∞. If it is multiplied by a function of higher order (or the same order) infinitesimal than it, then it is a bounded function
Otherwise, it's boundless

It is proved that the function y = 1 / (the square of x) is unbounded on (0,1)

It is necessary to prove that for any given positive value m, X belongs to (0,1) such that Y > M
In fact, if 01,1 / m

A simple calculation of 9 / 8 △ [(2 / 5, 1 / 10) X4 / 3]
9 / 8 ÷ [(2 / 5 + 1 / 10) X4 / 3]

9 / 8 ÷ [(2 / 5 + 1 / 10) X4 / 3]
=9 / 8 ÷ [1 / 2 x 3 / 4]
=9 / 8 ／ 3 / 8
=3

Point P1 (x1, Y1), point P2 (X2, Y2) , point PN (xn, yn) in function (x > 0)
On the images of △ p1oa1, △ p2a1a2, △ p3a2a3 All △ pnan-1an are equilateral triangles, with hypotenuse OA1, A1A2, a2a3 , an-1an are all on the x-axis (n is a positive integer greater than or equal to 2), then the coordinates of point P3 are (,); the coordinates of point PN are (,) (expressed by the formula containing n)

There is a black hill in the lower left corner of the display screen, 1cm ᦇ 178; the size of it has increased a small one these days

How many times the volume of Jupiter is that of the earth?
one thousand
one thousand and three hundred
three thousand
twelve thousand and ten

Jupiter is more than 1300 times the size of the earth, and is the largest planet in the solar system. Its four brightest moons are called Europa, Europa, Ganymede and Callisto for short, and their sizes are similar to the moon of the earth

If the function f (x) = XX2 + 2 (a + 2) x + 3a, (x ≥ 1) can use the mean value theorem to find the maximum value, then it is necessary to add that the value range of a is______ ．

∵ f (x) = XX2 + 2 (a + 2) x + 3A = LX + 3ax + 2 (a + 2) (x ≥ 1), if the function f (x) = XX2 + 2 (a + 2) x + 3a, (x ≥ 1) can use the mean value theorem to get the maximum value, the condition a satisfies is the condition g (x) = x + 3ax (x ≥ 1) can use the mean value theorem to get the minimum value. Obviously, a > 0, from x + 3ax ≥ 23a, if and only if x = 3ax, that is x = 3a, take "="; ∵ x ≥ 1 ∵ 3a ≥ 1, ∵ a ≥ 13 A ≥ 13

SiNx = 4 / 5, X is the value of acute angle SiNx / 2 plus or minus cosx / 2

(SiNx / 2 plus minus cosx / 2) ^ 2 = SiNx / 2 ^ 2 + cosx / 2 ^ 2 + 2 * SiNx / 2 * cosx / 2 = 1 + 2 * SiNx / 2 * cosx / 2 = 1 + SiNx = 9 / 5
X is an acute angle
x/2

The side length of a large square is 8 decimeters. Divide it into two rectangles of the same size and find out the perimeter of one of them
What is the length and width of the divided rectangle? Write the calculation process

Split right in the middle
The length of the rectangle: 8, width: 8 △ 2 = 4
Perimeter of rectangle: (8 + 4) × 2 = 24 (decimeter)

Lesson 3 Reading
People call a mountain across the whole basin () for two reasons: 1, () and 2 ()

Hengshan, because it's horizontal

Finding the general solution of the equation dy / DX = Y / (x + y ^ 3)

dy/dx=y/(x+y^3)
dx/dy=x/y+y²
Namely
dx/dy-1/y ·x=y²
therefore
x=e^[-∫（-1/y）dy] (∫y²e^[∫（-1/y）dy]dy+c)
=y (∫y²/y dy+c)
=y(∫ydy+c)
=y(y²/2+c)
=cy+1/2 y³