How to prove the limit of monotone bounded sequence

How to prove the limit of monotone bounded sequence

Let {x [n]} be monotonically bounded (let alone increasing), then there exists m > = x [n] (any n)
So {x [n]} has supremum, denoted as L
For any positive number a, there is a natural number n such that x [n] > l-a
Because x [n] is simple increasing, when n > = n, L-A

Limit problem of high number sequence!
The definition is: for any given positive number ε, there exists a positive integer n, such that when n > N,
|An-u|

You don't understand this definition yet. ε is arbitrary, so of course you can take a number greater than 1. The key to this definition is that n can be found for any given ε. Therefore, the smaller the value of ε, the more severe the condition is. However, no matter how small the value of ε is, such n can still be found. When n > N, | an-u|

We know that | x | = 8. | y | = 2

Because XY is less than 0
So x, y are positive and negative
Because | x | = 8. | y | = 2
So x = 8, y = - 2 or x = - 8, y = 2
XY=-16

Let f (XY, x + y) = x ^ 2 + y ^ 2 + XY, find f (x, y) / & 8706; X and f (x, y) / & 8706; y

The difficulty lies in the understanding of F (XY, x + y) = x ^ 2 + y ^ 2 + xy and variable substitution. X ^ 2 + y ^ 2 + xy = (x + y) ^ 2-xy = f (XY, x + y) replace the above formula with variable. V = XY. U = x + y. then f (U, V) = u ^ 2-v. if you don't like it, it's f (x, y) = y ^ 2-x. next, you can do it

Let E & # 178; - x + y + 2 = 0, make sure that z is a function of X and y, and find out the functions of &; Z / &; X and Z / &; y
Complete the steps

Is the Z power of E
Prototypic transformation
e²=x-y-2
Logarithm on both sides
Z=ln(x-y-2)
∂z/∂x=1/(x-y-2)
∂z/∂y=-1/(x-y-2)

G (x) = 0-x f (U) Du, where f (x) = 1 / 2 (X & # 178; + 1) 0 ≤ x < 1 1 / 3 (x-1) 1 to 2
Whether g (x) is continuous on interval 0-2

When 0 ≤ x

Let u = f (X & # 178; + Y & # 178;) and prove that y is partial Z / partial x-x, partial Z / partial y = 0

　　y(Du/Dx)-x(Du/Dy) = y*f'*(2x) - x*f'*(2y) = 0.

Given that the line y = KX + B is parallel to the line y = 3x-2 and passes through points (6,4), try to determine the analytic expression of the function,

y=3x-14
Because parallel, so the slope is the same, k = 3, so y = 3x + B, substituting x = 6, y = 4 into 4 = 3 * 6 + B, the solution is b = - 14, so the analytic formula of the function is y = 3x-14

If the analytic expression of Y with respect to X is y = kx-9, when x = 3, the function value is 6, then when x = - 3, the function value is ()
A.－6 B.－12 C.－24 D.－36

Choose C
6=3k-9
k=5
When x = - 3, y = 5 * (- 3) - 9 = - 24

The real part of analytic function f (x) = u (x, y) + IV (x, y) U (x, y) = (x ^ 2) - (y ^ 2) + 1 to find f (x) = u (x, y) + IV (x, y) satisfying the condition f (I) = 0

If f (I) = 0, then u (x, y) = u (I, y) = 0 and V (x, y) = V (I, y) = 0
From u (x, y) = (x ^ 2) - (y ^ 2) + 1, u (I, y) = I ^ 2-y ^ 2 + 1 = 0
∴y^2=i^2+1=0,y=0
Then when f (I) = 0, f (x) = u (I, 0) + IV (I, 0)