LIM (X -- x0) f (x) = 6, then f (x) at x0, a, must be continuous B, there must be f (x0) = 6 C, and there are left and right limits D And what's the concept of X tends to x0

LIM (X -- x0) f (x) = 6, then f (x) at x0, a, must be continuous B, there must be f (x0) = 6 C, and there are left and right limits D And what's the concept of X tends to x0

LIM (X -- x0) f (x) = 6 means that the limit exists when x tends to x0. Of course, there are left and right limits
c. There are left and right limits
Look at the definition of limit

If (LIM [f (x0 + 2 △ x) - f (x0)] / 3 △ x) = 1, then the value of F ′ (x0) is
A.1/2 B.3/2 C.1 D.2

lim[f(x0+2△x)-f(x0)]/3△x=1
So Lim [f (x0 + 2 △ x) - f (x0)] / △ x = 3
So Lim [f (x0 + 2 △ x) - f (x0)] / 2 △ x = 3 / 2
So f '(x0) = 3 / 2
Choose B

Let f (x) be differentiable at x = x0, find the limit LIM (XF (x) - x0f (x)) / (x-x0), and x approaches XO
Let f (x) be differentiable at x = x0, find the limit LIM (XF (XO) - x0f (x)) / (x-x0), and x approaches XO
wrote it wrong

Are you sure the title is right?
If you're right, the answer is f (x)
After f (x) is put forward by the molecule, the denominator x-x0 is approximately dropped,
For (XF (XO) - x0f (x)) / (x-x0), it's OK to use lobida's criterion directly
f(xo)-x0f'(x0))

If f '(x0) exists, find LIM (△ x-0) f ^ 3 (x0 + △ x) - f ^ 3 (x0 - △ x) / △ X=

As a result of the following formula = LIM (\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x0) * [F & # 1

If f '(x0) = 2, Lim [f (x0-k) - f (x0)] / 2K

Let H = - K, then LIM (K → 0) [f (x0-k) - f (x0)] / (2k) = - 1 / 2 × LIM (H → 0) [f (x0 + H) - f (x0)] / h = - 1 / 2 × f '(x0) = - 1

Given f ′ (x0) = - 2, find the value of LIM [f (x0-1 / 2K) - f (x0)] / K

lim [ f（x0-1/2k）-f（x0）]/[（x0-1/2k)-x0]=f′（x0）=-2
lim [ f（x0-1/2k）-f（x0）]·(-2k)=-2
lim [ f（x0-1/2k）-f（x0）]·(k)=1
(k tends to infinity)
You copied the title wrong

Lim △ x → 0 f (x0 + △ x) - f (x0) / △ x = k, then Lim △ x → 0 f (x0 + 2 △ x) - f (x0) / △ x = 2K, why is it online
lim △x→0 f（X0+2△X）—f （X0）/△X
=(lim △x→0 f（X0+2△X）—f （X0）/2△X)*2
Do variable substitution DT = 2DX
Because DX - > 0, DT - > 0
=(lim dt→0 f（X0+dt）—f （X0）/dt)*2
Let DT = △ X
There is
2*lim △x→0 f（X0+△X）—f （X0）/△X =2k
I don't understand the answer

This is the meaning of vector Lim △ x → 0 f (x0 + △ x) - f (x0) / △ x = vector

When x approaches to 1, the limit value of (x ^ m-1) / (x ^ n-1) is obtained,
N are all positive integers

By using the sum of the first n terms of the equal ratio sequence, we can get: 1 + X + x ^ 2 + x ^ 3 +. + x ^ (n-1) = (1-x ^ n) / (1-x) (1-x) [1 + X + x ^ 2 + x ^ 3 +. + x ^ (n-1)] = (1-x ^ n) or (x-1) [1 + X + x ^ 2 + x ^ 3 +. + x ^ (n-1)] = (x ^ n-1) and (x-1) [1 + X + x ^ 2 + x ^ 3 +. + x ^ (m-1)] = (x ^ m-1) / (x ^ n-1) = [1 + x ^

Limit of higher number: (a ^ x-1) / X what is the limit when x approaches 0?
Er, I haven't learned it. I only learned the algorithm of limit, the limit of E and lim (SiNx / x)

Should we learn the equivalent infinitesimal?
Let me try to solve it. For convenience, I use * instead of power
First, a * x is written as e * xlna, and then the molecule e * xlna-1 is replaced by its equivalent infinitesimal xlna
lim (a*x-1)/x =lim (e* xlna-1)/x=lim xlna/x = lna
I don't know if the answer is right

When x → x0, the limit does not exist. Is there only f (x) →∞ and the function undefined at x0

No. the strict definition of LIM (x → x0) f (x) nonexistence is: "at least one of F (x0-0) and f (x0 + 0) does not exist, or both of them exist but are not equal", which has nothing to do with whether the function has no definition at x0