The 1 / N power of LIM (n - > ∞) n I can't make it. I'll make it

The 1 / N power of LIM (n - > ∞) n I can't make it. I'll make it

The limit is equal to one
Let n ^ (1 / N) = 1 + t
Then n = (1 + T) ^ n
So n > 1 + [n (n-1) / 2] t ^ 2
Get t

How to calculate the N + 3 power (n →∞) of LIM (1 + 3 / N),

lim【x→∞】 (1+3/n)^(n+3)
=lim【x→∞】 [(1+3/n)^(n/3)]³×(1+3/n)³
=e³×1
=e³
Answer: e & # 179;

① Calculation: LIM ((C (n, 3)) / (the third power of N + 1)) =? (n - > 0) (where C is permutation)

C（n,3）=n!/[(n-3)!* 3!]
=(n-2)*(n-1)*n/6
C（n,3）/(n^3 + 1）=(n-2)*(n-1)*n/[6*(n^3 + 1)]
Because Lim [6 * (n ^ 3 + 1)] = 6 ≠ 0
So Lim {(n-2) * (n-1) * n / [6 * (n ^ 3 + 1)]}
= lim[（n-2)*(n-1)*n]/lim[6*(n^3 + 1)]
=0/6
=0

Given that x ∈ R, a = x2 + 12, B = 2-x, C = x2-x + 1, we try to prove that at least one of a, B, C is not less than 1

It is proved that if a, B and C are all less than 1, that is, a < 1, B < 1 and C < 1, then there is a + B + C < 3 and a + B + C = 2x2-2x + 12 + 3 = 2 (x − 12) 2 + 3 ≥ 3, which is contradictory; therefore, at least one of a, B and C is not less than 1

To the contrary, if 01 C (2-A) > 1, it seems that we can't get in touch,
Answer first
The first floor seems to be filled with water-
I don't see any connection.
Please answer the questions

So a (2-A) is less than or equal to 1 and a (2-B) > 1
No foul water
If a (2-B) > 1
Then B = a (2-A)
The same is true
b(2-c)>1 c1 a

Given the quadratic power of a - 3A + 1 = O, how much is the negative half of a + positive half of a

a²-3a+1=0
a²+1=3a
a+1/a=3
(a+1/a)²=3²
a²+2+1/a²=9
a²+1/a²=7
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F (x) for any real number x, y satisfies f (x + y) = f (x) + F (y). When x > 0, f (x)

Hehe ~ f (0 + 0) = f (0) + F (0) so f (0) = 2F (0) so f (0) = 0f (x + (- x)) = f (x) + F (- x) = f (0) = 0so f (x) + F (- x) = 0so f (x) = f (x) (so odd function) let X1 > x2 > 0f (x1-x2) = f (x1) + F (- x2) = f (x1) - f (x2) x1-x2 > 0so f (x1-x2)

In the range of real number, the following judgment is correct ()
A. If | x | = | y |, then x = Yb. If | x | = (y) 2, then x = YC. If x > y, then x2 > Y2D. If x > y, then x > y

A. When x and y are opposite to each other, the original formula does not hold, so this option is wrong; B. when x < 0, Y > 0, the original formula does not hold, so this option is wrong; C. when x > 0, y < 0 and | x | y | the original formula does not hold, so this option is wrong; D. because x > y, so x > y, so this option is correct

Let the real number x not equal to - 1, and prove that x ^ 2-6x + 5 of x ^ 2 + 2x + 1 is greater than or equal to the negative third

By means of analysis,
To prove that x ^ 2 + 2x + 1 / x ^ 2-6x + 5 is greater than or equal to minus one third,
Just prove x ^ 2-6x + 5 ≥ (x ^ 2 + 2x + 1) / 3,
That is to say, 3 (x ^ 2-6x + 5) + (x ^ 2 + 2x + 1) ≥ 0,
4X ^ 2-16x + 16 ≥ 0,
That is, 4 (X-2) ^ 2 ≥ 0 is obviously true,
So the original inequality holds

The function f (x) defined on R satisfies that for any real number m, N, there is always f (M + n) = f (m) × f (n)
And when x > 0, 0

Let m = 0, n > 0
0
f(0)=1
Let m + n = 0
f(0)=f(m)*f(-m)
=>
f(-m) = 1/f(m)
So when x1
For any x1, X2 belongs to R
x10,0
f(x2)/f(x1) = f(x2-x1)
f(x1)>f(x2)
This is a monotone decreasing function
1 let m > 0 > n
f(m+n)=f(m)f(n)
So f (M + n) / F (n) = f (m) < 1
And M + n ＞ n
So f (M + n) < f (n)
Let m > n > 0
f(m+n)=f(m)f(n)
So f (M + n) / F (n) = f (m) < 1
And M + n ＞ n
So f (M + n) < f (n)
The synthetic 1 2 F (x) decreases on R