Finding the limit limx approaching infinity (x ^ 2-1) / (3x ^ 2-x-1)

Finding the limit limx approaching infinity (x ^ 2-1) / (3x ^ 2-x-1)

Limx - > infinite [x ^ 2 + 1] / [x + 1] - ax-b = 2 find the value of a and B

The original formula is limx - > infinity [x ^ 2 + 1 - (x + 1) (AX + b)] / (x + 1) = limx - > infinity [(1-A ^ 2) x ^ 2 - (B + a) X-B] / (x + 1) = limx - > infinity [(1-A ^ 2) 2x - (B + a)] (lobita's law). Since X - > infinity, there must be 1-A ^ 2 = 0, (B + a) = 2. The solution is a = 1, B = - 3 or a = - 1, B = - 1

Given limx = > 1 [(x ^ 2 + ax + b) / (1-x)] = 1, find the value of a and B

Because x ^ 2 + ax + B is continuous at x = 1, it is bounded in the neighborhood of x = 1
And (limx --- > 1) [1-x] = 0
So (limx -- > 1) [x ^ 2 + ax + b] = 1 + A + B = 0
For its use, lobita has been improved
(limx----->1)[2x+a]/(-1)=(limx----->1)[-2x-a]=-2-a=-3
We get a = - 3
So B = - 1-A = 2

(2011 Zhengzhou three module) number {an}, a3=2, a7=1, if {1an+1} is arithmetic progression, then a11= ().
A. 0B. 12C. 23D. 2

Let the tolerance of the sequence {1An + 1} be d ∵ in the sequence {an}, A3 = 2, a7 = 1, and the sequence {1An + 1} is the arithmetic sequence ∵ 1A7 + 1 = 1A3 + 1 + 4D. Substitute A3 = 2, a7 = 1 into d = 124 ∵ 1a11 + 1 = 1A7 + 1 + 4D ∵ a11 = 12, so B is selected

In the arithmetic sequence. A2 + A5 = 11, A3 + A6 = 17, find the general term formula of sequence an

d=(17-11)/2=3
a2+a2+3d=11
2a2=2
a2=1
a1=a2-d=-2
an=-2+(n-1)3=3n-5

Given f (x + 1) = x2-4, in the arithmetic sequence {an}, A1 = f (x-1), A2 = - 32, A3 = f (x) (1) find the value of X and the general term formula an of the sequence {an}; (2) find A2 + A5 + A8 + +The value of A26

(1) This is the first time that we're going to get the a1 + a3 = 2A2, and the a1 + a3 = 2A2, and the a1 + a3 = 2A2, and the a1 + a3 = 2A2, and we're going to be (x + 1) 2-4 (x + 1) 2-4 (x + 1) (x + 1) (x + 1) (x + 1) = (x + 1) (x + 1-1) 2-2-4, and we're going to (x (x-1-1-1-1-1-1-1-1-1-1-1) 2-1) 2-2-2-2-4, and we're going to take these items from the series that we take out of these numbers from the series that are still equal difference numbers, and we're going to be the same difference numbers, and we're going to be the case we're going to be the case of the case that we are going to be the case of the case of the case of the A2 + A5 + A5 + A5 + A5 + A5 + it's not easy +A26 = 92 [− 32 − 32 (26 − 1)] = - 3512, when an = 32 (n − 3), A2 + A5 + +a26=92(−32−92+39)=2972．

The arithmetic sequence {an} has 2n + 1 terms, where a1 + a3 + +a2n+1=4，a2+a4+… +A2N = 3, then the value of n is ()
A. 3B. 5C. 7D. 9

The arithmetic sequence {an} has 2n + 1 terms, ∵ a1 + a3 + +a2n+1=4，a2+a4+… +A2N = 3, subtracting the two formulas, we can get a1 + nd = 1, adding the two formulas, we can get s2n + 1 = 7 = (2n + 1) a1 + (2n + 1) · 2n2d, ■ (2n + 1) (a1 + nd) = 7 ■ (2n + 1) = 7, ■ n = 3. So we choose a

In the arithmetic sequence (an), A1 = f (x-1), A2 = - 3 / 2, A3 = f (x)
(1) Find the value of X;
(2) Find the value of A2 + A5 + A8 + '+ A26!

(1)f(x+1)=x^2-2
Let x + 1 = t, x = T-1
f(t)=(t-1)^2-2
That is, f (x) = (x-1) ^ 2-2
a1=f(x-1)=(x-2)^2-2
a3=f(x)=(x-1)^2-2
Arithmetic sequence {an}
So a1 + a3 = - 2A2 = - 3
2x^2-6x+4=0
x^3-3x+2=0
x=1 or x=2
(2)
When x = 1, A1 = f (0) = - 1, A3 = f (1) = - 2
The tolerance is - 1 / 2
General term formula an = - 1-1 / 2 (n-1) = - (n + 1) / 2 = - 1 / 2 * (n + 1)
a2+a5+a8+``````+a26
=-1/2(3+6+9+.27)
=[-1/2][(3+27)*9/2]
=-135/2
When x = 2, A1 = f (1) = - 2, A3 = f (2) = - 1
The tolerance is 1 / 2
General term formula an = - 2 + 1 / 2 (n-1) = (N-5) / 2 = 1 / 2 * (N-5)
a2+a5+a8+``````+a26
=1/2(-3+0+3+.+21)
=[1/2][(-3+21)*9/2]
=81/2

It is known that in the arithmetic sequence an, A5 = 8, A10 = 18, three points (a1,0), (a2,2), (a3,0) are on the circle C
(1) Solving the equation of circle C
(2) If the chord length of the line L: MX + NY + 1 = 0 cut by the circle C is 2 √ 3, find the minimum value of M & # 178; + n & # 178
(3) If a moving straight line intersects circle C at two points a and B, and there is always | OA | * | ob | = 8 (point O is the origin of coordinates), try to explore whether the line AB is tangent to a fixed circle, please explain the reason

(1) So A1 = a5-4d = 0, A2 = 2, A3 = 4
The three points are (0,0), (2,2), (4,0)
Through the Pythagorean theorem, we know that these three points just form a right triangle. It is easy to know that the center of the circle is (2,0) and the radius is 2
So the equation of circle C is (X-2) ^ 2 + y ^ 2 = 4
(2) The chord length of line L cut by circle C is 2 √ 3. It is easy to find that the distance from the center of circle C (2,0) to line L is 1
That is to say, the line L is the tangent of the circle D: (X-2) ^ 2 + y ^ 2 = 1, and any tangent of a point (x0, Y0) on the circle D is taken. The equation is
（x0-2)(x-2)+y0y=1
It is concluded that (x0-2) x + y0y + 3-2x0 = 0
M = (x0-2) / (3-2x0), n = Y0 / (3-2x0) into m ^ 2 + n ^ 2 = [(x0-2) ^ 2 + Y0 ^ 2] / (3-2x0) ^ 2 = 1 / (3-2x0) ^ 2
We know that 1 ≤ x0 ≤ 3 (point on circle d)
So the minimum value of m ^ 2 + n ^ 2 is 1 / 9 (when x0 = 3)
(3) Let a (XA, ya) and B (XB, Yb) be on the circle C and satisfy | OA | * | ob | = 8
Namely √ XA ^ 2 + XB ^ 2 × √ XB ^ 2 + Yb ^ 2 = 8
(xA^2+xB^2) × (xB^2+yB^2)=64 ①
Because AB is on the circle C, so satisfy (X-2) ^ 2 + y ^ 2 = 4, substitute (XA, ya), (XB, Yb) into
It can be obtained that XA ^ 2 + Ya ^ 2 = 4xa, XB ^ 2 + Yb ^ 2 = 4xb, which can be substituted into formula (1)
We get xaxb = 4
Let the equation of line AB be MX + NY + 1 = 0, intersect with circle C, and the simultaneous equations are obtained
（m^2+n^2)x^2+(2m-4n^2)x+1=0,
A. B is the intersection
So xaxb = 1 / (m ^ 2 + n ^ 2) = 4, m ^ 2 + n ^ 2 = 1 / 4
It is easy to see that the distance from the point (0,0) to the straight line AB: MX + NY + 1 = 0 is a fixed value, and d = 2
The line AB is tangent to the definite circle x ^ 2 + y ^ 2 = 4

In the arithmetic sequence {an}, A2 + A8 = 16, A3 * A7 = 48, find the general term formula of the sequence, and explain that when d < 0, the sum of the first several terms is the largest?
Seeking process

Let the first item be A1 and the tolerance be d
Then:
a2=a1+d,a8=a1+7d
a3=a1+2d,a7=a1+6d
From the question: (a1 + D) + (a1 + 7D) = 16
(a1+2d)(a1+6d)=48
The solution is: A1 = 0, d = 2 or A1 = 16, d = - 2
When d = 0
We can get: n