LIM (x tends to 0) arctan2x / sin3x

2/3

LIM (x tends to 0) arcsin2x / 5x

x-->0 arcsin2x-->2x
LIM (x tends to 0) arcsin2x / 5x = 2 / 5

Lim {sin3x + XF (x)} / x ^ 3 (x tends to 0) find f (0), f '(0), f' (0)

There should be no mistake in finding the values of F (0) and f '(0) in the above formula, but f "(0) can't be found in that way
Now I take the original formula = Lim [3cos3x + F (x) + XF '(x)] / 3x ^ 2 = 0, and continue to use the law of Robida to get: Lim [- 9sin3x + 2F' (x) + XF "(x)] / 6x = 0. After simplifying, I get: F" (0) = 27, which seems to be inconsistent with your answer!
After this simplification, we can find that 3F "(0) = 27, f '(0) is 9
The last word above is wrong

Let Δ y = f (x0 + Δ x) - f (x0) and f (x) be differentiable at x = x0, then there must be ()
A.limΔy=0 B.Δy=0 C.dy=0 D.Δy=dy
Δx→0
=. = please give the analysis process

A.
Because it is differentiable at x0, Δ Y / Δ X has a limit when Δ X - > 0. Therefore, the limit of Δ y must be 0. Otherwise, the limit of Δ Y / Δ x is infinite and not differentiable

Let f (x) be differentiable at x0, then (F & # 178; (x) - F & # 178; (x0) / (x-x0) is the limit when x → x0

lim（f²(x)-f²(x0)/（x-x0)
The factorization is as follows
=lim (f(x)+f(x0))(f(x)-f(x0))/(x-x0)
Split into two
=lim[(f(x)+f(x0)] * lim [f(x)-f(x0)]/(x-x0)
According to the definition of derivative
=2f(x0)*f'(x0)

Given that f (x) is differentiable at x = a, (1) and f '(a) = B, then (1) limh → 0 [f (a + 3H) - f (A-H)] / 2H
By the way, help me solve (2) limh → 0 [f (a + H ^ 2) - f (a)] / h

1. Lim [f (a + 3H) - f (A-H) / 2H] = Lim {[3F '(a + 3H) + F' (A-H)] / 2} / / / the derivation of lobita's rule for H = Lim {[3F '(a) + F' (a)] / 2} = 2F '(a) = 2B2. Lim [f (a + H & # 178;) - f (a) / h] = Lim [3HF' (a + H & # 178;)] / / / the derivation of lobita's rule for H = 0

Derivative problem Lim [f (a + H ^ 2) - f (a)] / h =?
（1）lim [f(a+h^2)-f(a)]/h=?
（2）lim [f(a+3h)-f(a-h)]/2h=?

lim [f(a+h^2)-f(a)]/h=h*lim [f(a+h^2)-f(a)]/h^2=h*f'(a);
lim [f(a+3h)-f(a-h)]/2h=2*lim [f(a+3h)-f(a-h)]/(2h*2)= 2*f'(a);

Is the derivative of a continuous function continuous?

Not necessarily
(1) There are many examples of continuous derivatives of continuous functions, such as
F (x) = x, f '(x) = 1, obviously f' (x) is continuous in (- ∞, + ∞)
(2) Examples of discontinuous derivatives of continuous functions
f(x)= x²sin(1/x) (x≠0)
0 (x=0)
f'(0)=lim(x→0)[f(x)-f(0)]/(x-0)=lim(x→0)[xsin(1/x)]=0
∴f'(x)= 2xsin(1/x) -cos(1/x) (x≠0)
=0 (x=0)
F '(x) is discontinuous at x = 0

Take an example where the function is continuous but the directional derivative does not exist

Z = under the root sign (x ^ 2 + y ^ 2) is continuous at (0,0), but the directional derivative in any direction does not exist, because on both sides there is a decreasing velocity of one and an increasing velocity of one. This is similar to the derivability of | x | at 0

Who can give an example to show that the original function is differentiable, but its derivative is not necessarily continuous, and give an image
You have to bring images,

If f (x) = x ^ 2 * sin (1 / x), and f (0) is defined as 0, then f (x) is differentiable (when x is not zero, it is obviously differentiable. At x = 0, there is a definition, derivative, derivative is 0), but the derivative of F (x) is discontinuous at x = 0! Its derivative is - cos (1 / x) + 2 * x * sin (1 / x), the latter part is continuous at x = 0, but the former part is continuous at x = 0

LIM (x tends to 0) arctan2x / sin3x