Let f (0) = 0 and f '(0) exist, then limx tends to be the same as 0f (x) / X=

Let f (0) = 0 and f '(0) exist, then limx tends to be the same as 0f (x) / X=

When limx tends to zero, limf (x) / x = f '(0)

If f (x) is continuous at x = 0 and limx → 0f (x) / X exists, it is proved that f (x) is differentiable at x = 0

Because f (x) is continuous at x = 0 and limx → 0, f (x) / X exists
So f (0) = LIM (X -- > 0) f (x)
= lim (x-->0) f(x)/x * x = lim (x-->0) f(x)/x * lim (x-->0) x = 0
So: let limx → 0, f (x) / x = a
lim (x-->0) |（f(x) - f(0)) / (x -0) - A| = lim (x-->0) |f(x) / x - A| = | lim (x-->0) f(x) / x - A | = 0
That is, f '(0) = a exists

limx→∞(1+1/2x)^3x+2

limx→∞(1+1/2x)^3x+2
=
limx→∞(1+1/2x)^2x*(3x+2)/(2x)
=e^limx→∞(3x+2)/(2x)
=e^(3/2)

In the arithmetic sequence {an}, A6 = 5, A3 + A8 = 5, find S9

A3 + A8 = A4 + A7 = A5 + A6 = 5. So A5 = 0. S9 = 5a5 = 0

In the arithmetic sequence {an}, A6 = A3 + A8, then S9=______ ．

Let the tolerance of {an} be D, and the first term be A1. From the meaning of the question, we can get that a1 + 5D = a1 + 2D + A1 + 7d, ｜ a1 + 4D = 0, S9 = 9a1 + 9 × 82d = 9 (a1 + 4D) = 0, so the answer is 0

In known arithmetic sequence, A6 = 5, A3 + A8 = 5, find the first term A1 and tolerance D

Because: A6 = 5, A3 + A8 = 5
So: a1 + 5D = 5, a1 + 2D + A1 + 7d = 5
The solution is: A1 = - 20, d = 5

The arithmetic sequence, A6 = A3 + A8, then a1 + A9 =?
How to calculate

D is the tolerance of arithmetic sequence, A6 = a1 + 5D, A3 = a1 + 2D, a8 = a1 + 7d, so (a1 + 5d) = (a1 + 2D) + (a1 + 7D), A1 = - 4D
So a1 + A9 = a1 + (a1 + 8D), substituting A1 = - 4D, a1 + A9 = - 4D-4D + 8D = 0

In the sequence {an}, A3 = 2, a7 = 1, and the sequence {1 / (an + 1)} is an arithmetic sequence, then an =?

Let BN = 1 / (an + 1), then BN is an arithmetic sequence, and the tolerance is d
b3=b1+2d=1/3,b7=b1+6d=1/2
So d = 1 / 24, B1 = 1 / 4
bn=1/24+(n-1)/4=(n+5)/24
That is 1 / (an + 1) = (n + 5) / 24
an=(19-n)/(n+5)

In the sequence {an}, A3 = 2. A7 = 1, and the sequence {1 / an + 1} is an arithmetic sequence, then a11 is equal to
Let BN = 1 / (an + 1)
Then B3 = 1 / 3
b7=1/2
So the tolerance d = (1 / 2-1 / 3) / 4 = 1 / 24
So B11 = 1 / 2 + 1 / 24 = 13 / 24 = 1 / (a11 + 1)
So a11 = 11 / 13
Right

Step five starts to have problems
Let BN = 1 / (an + 1)
Then B3 = 1 / 3
b7=1/2
So the tolerance d = (1 / 2-1 / 3) / 4 = 1 / 24
Therefore, B11 = A7 + 4D = 1 / 2 + 4 * 1 / 24 = 2 / 3 = 1 / (a11 + 1)
Dissolve 2 / 3 = 1 / (a11 + 1) to get: a11 = 1 / 2
The above is solved according to your idea, or by the equal difference neutral method
Let BN = 1 / (an + 1)
Then B3 = 1 / 3
b7=1/2
It can be seen from the equal difference neutral that:
b3+b11=2*b7
Then, B11 = 2 / 3 = 1 / (a11 + 1)
Dissolve 2 / 3 = 1 / (a11 + 1) to get: a11 = 1 / 2

In the sequence {an}, A3 = 2, a7 = 1, if {an + 1 / 1} is an arithmetic sequence, then a11=（

Let 1 / (an + 1) = bna3 = 2, a7 = 7, then 1 / (A3 + 1) = 1 / 3, 1 / (A7 + 1) = 1 / 2, that is, B3 = 1 / 3, B7 = 1 / 2, because BN is an arithmetic sequence, the tolerance of BN is d = (b7-b3) / 4 = 1 / 24, so the general term of BN is BN = 1 / 3 + (n-3) XD = (n + 5) / 24, so B11 = 2 / 3, so the solution of equation 1 / (a11 + 1) = B11 = 2 / 3 is obtained