Limx (√ x ^ 2 + 1-x) x tends to zero and positive infinity

Limx (√ x ^ 2 + 1-x) x tends to zero and positive infinity

Multiplication √ (X & # 178; + 1) + X
Molecular square difference = x & # 178; + 1-x & # 178; = 1
So the original formula = limx / [√ (X & # 178; + 1) + x]
Divide up and down by X
=lim1/[√(1+1/x²)+1]
=1/2

Find limx approaching 0 [ln (1 + x) - x) / x ^ 2]

Type 0 / 0
Using the law of lobida
=lim[1/(1+x)-1]/2x
=lim[-1/(1+x)]
=-1

1. Find the n-order derivative of Xe ^ - 2x. 2. Find the n-order derivative of x ^ 2 + LNX in 50 minutes. Use Leibniz formula
Using Leibniz formula, 50 points to find the detailed method
1. Find the point x such that f ^ n (x) = 0
2. Find the value range of F '' (x) > 0, X

1、y=xe^(-2x)
There are only two terms in this n-order derivative, one is e ^ (- 2x) to obtain n-order derivative, X does not; the other is e ^ (- 2x) to obtain n-1 order derivative, X to obtain first order derivative, and the other terms are all 0 because the order of X is ≥ 2
y^(n)=x[e^(-2x)]^(n)+C(50,1)(x)'[e^(-2x)]^(n-1)
=(-1)ⁿ2ⁿxe^(-2x)+(-1)ⁿ⁻¹n*2ⁿ⁻¹e^(-2x)
=(-1)ⁿ2ⁿ⁻¹e^(-2x)(2x-n)
2、y=x²+lnx
This problem does not need Leibniz formula
y'=2x+1/x
y''=2-1/x²
y'''=2/x³
.
Y ^ (n) = (- 1) & ᦉ 8319; & ᦉ 8315; & ᦉ 185; (n-1)! / X & ᦉ 8319;, when n ≥ 3
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A mathematical examination of Leibniz formula of n-order derivative of product function?
For example, the problem is to find the derivative of order n, not other Leibniz formula,

The HKCEE, however, is flexible
For example: F (x) = a (x) * B (x)
Where B (x) is a quadratic trinomial, then finding the third derivative becomes 0
In fact, there are only the first three terms in Leibniz expansion
This is the basic type of debut topic

Find the derivative of order 10: y = x ^ 10 / (1-x) Leibniz formula, is there any good method?
/(1-x) ^ 11 / x0c the answer is just the sum of the fractions and the denominator of the fraction. It's a coincidence. What skills should we have

y=x^10/(1-x)
=y=（x^10-1+1）/(1-x)
=-（x^9+x^8+...+x+1）+1/(1-x)
Because the previous polynomial - (x ^ 9 + x ^ 8 +... + X + 1) has the highest power of 9, so
10 order derivative = 0
therefore
Just calculate the 10 order derivative of 1 / (1-x)
There is a formula,
y'=1/(1-x)²
y''=2!/(1-x)³
.
therefore
y^(10)= 10!/(1-x)^11

Sum: 1 + 2x + 3x ^ 2 + +nx^n-1

When x = 1, Sn = 1 + 2 + 3 +... + n = n (n + 1) / 2x ≠ 1, Sn = 1 + 2x + 3x ^ 2 +... + NX ^ (n-1) xsn = x + 2x ^ 2 +... + (n-1) x ^ (n-1) + nx ^ NSN xsn = (1-x) Sn = 1 + X + x ^ 2 +... + x ^ (n-1) - NX ^ n = (1-x ^ n) / (1-x) - NX ^ NSN = (1-x) ^ 2 - NX ^ n / (1-x)

Sum 1 + 2x + 3x & # 178; + +Nx ^ n-1 (n-1 is index)

Sn=1+2x+3x^2+… +Nx ^ n-1xsn = x + 2x ^ 2 + 3x ^ 3 +... + (n-1) x ^ (n-1) + NX ^ n (1-x) Sn = (x + x ^ 2 +... + x ^ n-1) + 1-nx ^ n = x (x ^ n-2) / X-1 + 1-nx ^ n. in addition, the case of x = 1, Sn = 1 + 2 + 3 +... + n = n (n + 1) / 2 is discussed separately

Sum 1 / 3 + 2x (1 / 3 ^ 2) + 3x (1 / 3 ^ 3) + +nx(1/3^n)
1/3+2x1/3^2+3x1/3^3+…… +NX1 / 3 ^ n sum SN

sn=1/3+2x1/3^2+3x1/3^3+…… +nx1/3^n;...⑴
3/1sn=1/3^2+2x1/3^3+3x1/3^4+…… +nx1/3^n+1;.⑵
(1)-(2)=2/3sn=1/3+(1/3^2+…… +1/3^n)-nx1/3^n+1;
Simplification is SN

Sum s = 1 + 2x + 3x ^ 2 +. + NX ^ (n-1) =?
The power of X is increasing, the coefficient is a natural number, and the sum is obtained

Because s = 1 + 2x + 3x ^ 2 + + nx^(n-1) …… In duplicate
So XS = x + 2x ^ 2 + 3x ^ 3 + + nx^n …… Two forms
Subtract two from one
( 1-x )S = 1 + x + x^2 + x^3 + … + x^(n-1) - nx^n
( 1-x )S = ( 1 - x^n )/( 1-x ) - nx^n
So s = (1 - x ^ n) / (1-x) ^ 2 - (NX ^ n) / (1-x)

Sum Sn = 1 + 2x + 3x ^ 2 +. Nx^
x+x^2+x^3+…… X ^ n = (x ^ (n + 1) - 1) / (x-1) - 1. The result is obtained by deriving the above formula

Note the equation: SN = 1 + 2x + 3x ^ 2 +... + NX ^ (n-1)
And xsn = x + 2x ^ 2 + 3x ^ 3 +... + NX ^ n
Sn-xSn=1+x+x^2+x^3+...+x^(n-1)-nx^n
The sum of x = 1 is very simple, X ≠ 1 is used to calculate the sum of the first n terms of the equal ratio sequence