If f (x) is continuous at x = x0, then. Limf (x) / (x-x0) = a if and only if f (x0) = 0, and the first derivative of F (x0) is a, how can we prove it?

Proband necessity:
From limf (x) / (x-x0) = a
Because the denominator tends to zero and the limit of the formula exists, the molecule should also tend to zero, and f (x) is continuous at x0
So limf (x) = 0 = f (x0)
So the limit can be written as LIM (f (x) - f (x0)) / (x-x0) = a
From the definition of derivative, we know the derivative f '(x0) = a at x0
The sufficiency of reexamination:
Because the first derivative of F (x0) is a
So LIM (f (x) - f (x0)) / (x-x0) = a
Because f (x0) = 0
So limf (x) / (x-x0) = a
It's over

The known sequence {an} satisfies: 1 * a1 + 2 * A2 + 3 * A3 +. + n * an = n (n of an is subscript) (1) find the sequence formula (2) if BN = 2 ^ n / an
Finding the sum of the first n terms of {BN}

1*a1+2*a2+3*a3+.+n*an=n
1*a1+2*a2+3*a3+.+(n-1)*a(n-1)=n-1
Subtraction of two formulas
n*an=1
an=1/n
bn=2^n/an
=2^n/(1/n)
=n*2^n
sn=1*2^1+2*2^2+3*2^3+.+n*2^n.1
2sn=1*2^2+2*2^3+3*2^4+.+n*2^(n+1).2
Formula 1-2
-sn=2^1+2^2+2^3+2^4+.+2^n-n*2^(n+1)
-sn=2*(1-2^n)/(1-2)-n*2^(n+1)
-sn=2^(n+1)-2-n*2^(n+1)
sn=n*2^(n+1)-2^(n+1)+2
sn=(n-1)*2^(n+1)+2

In the sequence {an}, A1 = 1, A2 = 4, and a (n) + a (n + 1) = 4N + 1 to find the general term formula, () is the subscript

an+a（n+1）=4n+1-------(1)
a(n-1)+an=4n-3------------(2)
a(n+1)-a(n-1)=4
1) When n is even, (n is greater than or equal to 2)
a3-a1=4
a5-a3=4
.
a(n+1)-a(n-1)=4
Superposition method
A (n + 1) - A1 = 2n is obtained
a(n+1)=2n+1
An = 2N-1 (n is odd)
2) When n is odd
a4-a2=4
a6-a4=4
.
a(n+1)-a(n-1)=4
Superposition method
We obtain a (n + 1) = 2n + 2
An = 2n (n is even)
So an = 2N-1 (n is odd)
An = 2n (n is even)

In the sequence {an}, A1 = 1, A2 = 4, and a (n) + a (n + 1) = 4N + 1, find the general term formula
an+a(n+1)=4n+1
a(n-1)+an=4n-3
Subtracting a (n + 1) - A (n-1) = 4
So A1, A3, A5 Is an arithmetic sequence with a tolerance of 4
A2, A4 and A6 are also arithmetic sequences with tolerance of 4
I agree with that
But why is it wrong for me to do this
a(n)+a（n-1）=4(n-1)+1=4n-3
a(n-1)+a(n-2)=4(n-2)+1=4n-7
a(n-2)+a(n-3)=4(n-3)+1=4n-11
.
a2+a1=41+1=5
It is an arithmetic sequence with tolerance of 4, and the sum is calculated with (first term + last term) number of terms / 2
An + A1 = (5 + 4n-3) * (n-1) / 2 is obtained by superposition method
So an = (2n + 1) * (n-1) = 2n ^ 2-n-1
I hope it can solve my problem

The superposition method is applicable to the subtraction of two adjacent terms, so that the middle phase can be eliminated. After the superposition, there are many terms in the middle, which cannot be eliminated. You mistakenly take it as a minus sign,

If the sequence {an} satisfies A1 = 2, an + 1 = 1 + an1 − an (n ∈ n *), then a1a2a3 The value of A2010 is ()
A. 1B. -3C. 2D. -6

∵1=2，an+1＝1+an1−an(n∈N*)，∴a2＝1+a11−a1=1+21−2=-3，a3＝1+a21−a2=1−31+3＝−12a4＝1+a31−a3=1−121+12=13a5＝1+a41−a4=1+131−13=2=a1∴a6=a2，a7=a3，… A2009 = A1, A2010 = A2 sequence {an} is a periodic sequence with period of 4, and a1a2a3a4 = 2 × (− 3) × (− 12) × 13 = 1  a1a2a3 A2010 = - 6, so select D

If the sequence {an} satisfies A1 = 1 and 1 / [a (n + 1)] - 1 / an = 1, then A1 * A2 + A2 * A3 + +a2010*a2011=?

1/[a(n+1)]-1/an=1
1/a1=1
So 1 / an is an arithmetic sequence with 1 as the first term and 1 as the tolerance
1/an=1+1(n-1)=n
an=1/n
a1*a2+a2*a3+… +a2010*a2011
=(1/1)(1/2)+(1/2)(1/3)+.+(1/2010)(1/2011)
=1/(1*2)+1/(2*3)+.+1/(2010*2011)
=1-1/2+1/2-1/3+.+1/2010-1/2011
=1-1/2011
=2010/2011
If you don't understand, you can ask
Please accept if you are satisfied
thank you

Given that the decreasing arithmetic sequence {an} satisfies A12 = a92, when the first n terms and Sn of the sequence {an} take the maximum value, n = ()
A. 3b. 4C. 4 or 5D. 5 or 6

∵ the decreasing arithmetic sequence {an} satisfies A12 = a92, ∵ A1 = - A9 > 0, i.e. A1 = - a1-8d; ∵ a1 + 4D = 0, i.e. A5 = 0; ∵ S4 = S5, at this time, the first n terms and Sn of the sequence {an} are the largest; therefore: C

Find LIM (Sn + Sn + 1) / (Sn + sn-1), an is arithmetic sequence, A1 is not zero

Let Sn = a1 + (n-1) d, then Sn = A1 * n + n (n-1) d / 2 is substituted into LIM (Sn + Sn + 1) / (Sn + sn-1) = [Na1 + n (n-1) d / 2 + (n + 1) * a1 + n ^ 2D] / [Na1 + n (n-1) d / 2 + (n-1) a1 + (n-2) (n-1) d / 2] and divided by n ^ 2 to get = Lim [1 / Na1 + (1-1 / N) d / 2 + (1 / N + 1 / N ^ 2) * a1 + D] / [1 / Na1 + (...)

In the arithmetic sequence {an}, A1 = 1 / 25, the tenth term begins to be more than 1, and lim (1 / N & sup2;) * (an + SN) = t?
SN is the sum of all the terms of which an
Which Lim is followed by the square of N in brackets

Let the tolerance be D, then A10 = a1 + 9D = 1 / 25 + 9D > 1D > 8 / 75t = LIM (1 / N & sup2;) * (an + SN) = Lim [an + (a1 + an) n / 2] / N & sup2; = Lim [2An + Nan + Na1] / (2n & sup2;) = Lim [n ^ 2D + (n-2) d + (2n + 2) A1] / (2n & sup2;) = D / 2 > 4 / 75t > 4 / 75

Let the sum of the first n terms of an be Sn, and A1 = 1, an + 1 = 1 / 3Sn,
Find the general term formula of sequence an

An+1=1/3Sn 3An+1=Sn(1) 3An=Sn-1(2)
(1) - (2) we get 3an + 1 = 4An (n is greater than or equal to 2), so an is an equal ratio sequence with A2 as the first term and q = 4 / 3
A2 = 1 / 3A1, so A2 equals 1 / 3
An = 1 / 3 * (4 / 3) ^ (n-2) (n ≥ 2)
A1 = 1 does not satisfy an
So an = 1 (n = 1)
=1 / 3 * (4 / 3) ^ (n-2) (n ≥ 2)

If f (x) is continuous at x = x0, then. Limf (x) / (x-x0) = a if and only if f (x0) = 0, and the first derivative of F (x0) is a, how can we prove it?