The limit of the request function ㏑ x * (3x-2x ^ 2) when x approaches 0,

The limit of the request function ㏑ x * (3x-2x ^ 2) when x approaches 0,

When X - > 0:
Original formula = LNX / (1 / (3x-2x ^ 2))
Using the law of lobita, the derivation of numerator and denominator is as follows
Original = - (3x-2x ^ 2) ^ 2 * (3-4x) / X
=-x*(3-2x)^2*(3-4x)
->0 (x->0)
So the limit is 0

(1) Limx tends to zero, x times sin1 / x, how much (2) tends to positive infinity

(1) The product of 0 and bounded function is infinitesimal, so the limit is 0
(2) When it approaches infinity, it is 1, which can be inferred from the first important limit

Limx tends to 0, xsin (1 / x) = 0, why is it equal to 0
If I change this formula to sin (1 / x) / (1 / x), isn't the first important limit equal to 1

You mean LIM0 > SiNx / x = 1
But this problem is Lim ∞ > sin (1 / x) / (1 / x), the two are different
-1/(1/x)≤sin(1/x)/(1/x)≤1/(1/x)
lim-1/(1/x)=0=lim1/(1/x)
The limit = 0 is obtained by the pinch theorem

The derivative of implicit function x ^ (1 / 2) + y ^ (1 / 2) = 9, and the slope of tangent at point (1,64),
In particular, if y ^ (1 / 2) is equal to y '/ [2 (y ^ (1 / 2))], I haven't learned yet

[x^(1/2)+y^(1/2)]'=0,1/[2(x^(1/2)]+y'/[2(y^(1/2))]=0,y'=-(y/x)^(1/2),y'(1)=-64^(1/2)=-8.
The tangent equation is y-64 = - 8 (x-1), 8x + y-72 = 0

What is the slope of the tangent of the function y = x ^ 3 when x = 1?
rt

Y '= 3x ^ 2, when x = 1, y' = 3, the slope of tangent is 3

Finding the original function of derivative (LNX) ^ 2

Answer: ∫ (LNX) ^ 2DX = x (LNX) ^ 2 - ∫ x * D ((LNX) ^ 2) = x (LNX) ^ 2 - ∫ x * 2lnx / xdx = x (LNX) ^ 2-2 ∫ lnxdx = x (LNX) ^ 2-2x * LNX + 2 ∫ XD (LNX) = x (LNX) ^ 2-2x * LNX + 2 ∫ D (x) = x (LNX) ^ 2-2x * LNX + 2x + C (C is any real number), so the original function of (LNX) ^ 2 is x (LNX) ^ 2-2x * LNX

Function y (x) = x ^ 2 * LNX, derivation

y(x)=x^2*lnx
y'=(x²)'*lnx+x²*(lnx)'
=2x*lnx+x²*(1/x)
= 2xlnx+x

A function is differentiable at one point. It is continuous at this point, so is its derivative continuous at this point? Let's talk about it

Not necessarily. A typical example is the function f (x) = x ^ 2 * sin (1 / x), X is not equal to 0; f (0) = 0. For this f (x), we can know that f '(0) = 0 by using the derivative definition at 0, but when x is not equal to 0, f' (x) = 2xsin (1 / x) - cos (1 / x), it is easy to see that f '(x) is not continuous at 0

If the function is differentiable and the derivative is continuous, is the derivative differentiable?

Not necessarily
For example, piecewise function
F (x) = x ^ 3 * sin (1 / x) x is not equal to 0
0 x=0

A function has continuous second derivative in a certain interval. What does this sentence mean

When the second derivative is more than 0, the interval is concave. When the second derivative is more than 0, the interval is concave