A problem on the limit of binary function Find the limit of a binary function f (x, y) when both X and Y tend to 0. If the limit exists, then it means that the limit approaching by any path is the same,

That's right
In order to prove the existence of limit, we can first prove that it is differentiable, differentiable must be continuous, continuous limit exists. But if it is not differentiable, limit may also exist

Request: limit binary function
lim (x2+y2 )1/2-sin(x2+y2 )1/2
(x,y→0) (x2+y2 )3/2
Please help to solve the problem, the numbers are superscript
(x2 + Y2) 1 / 2-sin (x2 + Y2) 1 / 2 divided by (x2 + Y2) 3 / 2
When XY tends to 0, we can find the limit of binary function

Let u = √ (x ^ 2 + y ^ 2), then when (x, y) → (0,0), u → 0, the problem is transformed into the limit of univariate function: LIM (U → 0) (u-sinu) / u ^ 3, and the result is 1 / 6 by the law of lobita

Limit definition of binary function
Why are the two conditions equal

The simple point that eposilon and delta said, the nonstandard point is the infinitesimal positive number,
X-x0 wireless tends to zero, such as x = 4, x0 = 4.00.001,
Isn't the top radical test a plane distance formula,

Find the number of nonnegative integer solutions of the multivariate linear infinitive X1 + x2 + X3 +... + xn = K
It's about higher algebra,

This question should be a question of permutation and combination in high school?
The main solution is "baffle method"
For example: m same balls into n boxes, each box at least one
M balls, M-1 gap; divided into N parts, n-1 baffle;
The result is C (n - 1, M - 1); you can test it with a few simple data
After understanding the above example, let's prove your proposition
First, change x1, X2,. Xn to positive integers
So let X1 = X1 + 1, X2 = x2 + 1,. Xn = xn + 1
That is, X1 + x2 +. + xn = n + K
Now you see, it's equivalent to putting n + K identical balls into n boxes
So the result is C (n-1, N + k-1) = C (k, N + k-1)

If data x1, X2 The average of XN is x, then (x1-x) + (X2 - x) + +Does (xn-x) = 0 always hold?

Yes
Because X1 + x2 +. + xn = NX
Open the brackets of the following expressions, and the order of rearrangement is X1 + x2 +. + xn NX = 0
So it's always true

The length of a line is measured n times, and N results x1., X2 Xn, if we use X as the approximate value of the length of this line, when x takes what value, (x-x1) 2 = (x-x2) 2 + +The value of (x-xn) 2min? X is related to which common statistic?

X-x1) ^ 2 + (x-x2) ^ 2 +... + (x-xn) ^ 2 can be reduced to NX ^ 2-2 (x1 + x2 +... Xn) x + X1 ^ 2 + x2 ^ 2. According to the properties of quadratic function, when x = - B / 2A = (x1 + x2 +... + xn) / N, the function is the smallest
What I just finished is related to the average

The root of the equation f (x) = x is called the fixed point of F (x). If f (x) = Xa (x + 2) has a unique fixed point and X1 = 1000, xn + 1 = 1F (& nbsp; 1xn) & nbsp; (n ∈ n *), then x2011=______ ．

From Xa (x + 2) = x, we get AX2 + (2a-1) x = 0. Because f (x) has a unique fixed point, 2a-1 = 0, that is, a = 12. So f (x) = 2XX + 2. So xn + 1 = 1F (& nbsp; 1xn) & nbsp; = 2xn + 12 = xn + 12. So x2011 = X1 + 12 × 2010 = 1000 + 20102 = 2005

Let x ^ n + x ^ (n-1) + x ^ (n-2) +. + x = 1 have a unique root xn in (0,1), and find LIM (n → + ∞) xn
The answer is 0 or 1 / 2.

Simple subtraction of proband sequence
We get the existence
And the polar root must not be equal to 1
Then use the example of equal ratio to get the sum, and the answer is 0

Given the sequence x0 = 1, the nth term of x = 1 / (the Nth-1 term of 1 + x), try to prove that the sequence {xn} converges

x=1/(1+x)
|x-x=|1/(1+x)-1/(1+x)|
=|(x-x)/[(1+x)(1+x)]|
And x > 0, so x + 1 > 1 is constant
So there exists Q > 0 such that
So 1 / | (1 + x) (1 + x)|

If S10 = 4s5, then A1: D equals ()
A. 14B. 12C. 2D. 4

Let A1 = a, S10 = 10A + 10 (10 − 1) D2 = 10A + 45d, S5 = 5A + 5 (5 − 1) D2 = 5A + 10d, from S10 = 4s5, 10a + 45d = 4 (5a + 10d), i.e., 10a + 45d = 20A + 40d, ∵ 5D = 10a, ∵ D ≠ 0, ∵ ad = 510 = 12, i.e., A1: D = 12

A problem on the limit of binary function Find the limit of a binary function f (x, y) when both X and Y tend to 0. If the limit exists, then it means that the limit approaching by any path is the same,