When limx → 0, [√ (1 + x) - 1] / x, it is better to have calculation process

When limx → 0, [√ (1 + x) - 1] / x, it is better to have calculation process

Let t = √ (1 + x), then x = T ^ 2-1
Original formula = (t-1) / (T ^ 2-1) = 1 / (T + 1)
Because limx → 0
So limt → 1
Lim = 1 / 2

Finding the limit of limx → 0 1 / X 1n (1 + x)
Please write down the steps to solve the problem

X tends to 0, and the equivalent infinitesimal of Ln (1 + x) is X
The original formula is (1 / x) x = 1
above
I answered first

Limx tends to infinity (1-2 / x) ^ X / 2-1 find its limit

[50 points calculus for advanced mathematics]
Let f (x) be continuous on [a, b], and in (a, b), f (a) f (b) > 0 f (a) f [(a + b) / 2]

Please see the picture for the detailed answers. If you have any questions, please contact me

1. Set function
F (x) = {x square of 1 / E, X ＜ 0
={a+X ,X >=0
The limit Lim f (x) exists when the constant a is sum
2. It is unnecessary to find out the derivative of function f (x) = x (X-2) (x-3). It is shown that f '(x) = 0 has several real roots, and the interval of each root is pointed out

1. How can we not understand this? It is obviously a piecewise function
F (x) is continuous at any point where x is not equal to 0, so the limit exists
At x = 0
f(0+0)=a
f(0-0)=1
If limf (x) exists, then a = 1
2. F (x) is a cubic function of X, and its derivative is a quadratic function of X, which has at most two real roots
f(0)=f(2)=f(3)=0
Rolle theorem
There exists a belonging to (0,2) such that f '(a) = 0
There exists B belonging to (2,3) such that f '(b) = 0
So there are two real roots in the interval (0,2), (2,3)

Higher number problem (calculus calculation)
How to get x (T) = X. (e ^ RT) from DX / dt = RX

dx/dt=rx
dx/x=rdt
Two side integral
Ln | x | = RT + C 'C' is a constant
X (T) = CE ^ RT C is a constant
x(0)=Ce^0=C
So x (T) = x (0) e ^ RT

Calculus in Advanced Mathematics
1. Let f (x) be a periodic function with period 2, and (piecewise function) f (x) = x, 0

I finished the answer to the second question and took a picture~
I don't quite understand the first question, sorry~

The position of calculus in Higher Mathematics

As the saying goes: if the brain wants to be flexible, it must learn mathematics. Mathematics can cultivate three kinds of thinking: one is reverse thinking, and the story of Sima Guang smashing the VAT is a classic example of reverse thinking; the other is divergent thinking, which is the ability cultivated by multiple solutions to one problem in mathematics; the third is systematic thinking, Any individual who lives in one or more relative systems must learn to look at problems in the system and from the overall situation and the whole. It is the so-called "pull one hair and move the whole body". In order to achieve harmony in the system, the members of the whole system must have a high degree of system consciousness and can not simply look at problems in isolation and one sidedly
There are two important calculations in Calculus: differential calculation and integral calculation. One is positive and the other is negative, which is a typical opposite operation in mathematics. Each of them is in a system. The emergence of calculus solves many practical problems and plays a strong foundation for the development of the whole science, especially for the contribution to physics. It can be said that without the birth of calculus, physics will be in the same place, The whole society will return to the 16th century. Therefore, all universities should set up the course of calculus, so that students can understand it, further establish scientific awareness, learn science, know science and use science, firmly establish the scientific outlook on development, and be a qualified contemporary college student

The following variables are infinitesimals in a given change process ()
A.2^(-X)-1
B.SINX/X
I don't have any idea about this topic, and I don't know the basis for judging infinitesimal
The X in option AB tends to zero

The answer is a
The most basic method is to find out the limit value, which is 0 is infinitesimal
The limit of B is 1

Two problems of calculus in Higher Mathematics
Z = x ^ y is partial derivative of Y, the value at (1 / x, 1) is DZ / dy │ (1 / x, 1) = ()
There is also a question z = cot (2XY ^ 2), then & #; Z / & #; y = ()

lnZ=ylnx
(әz/әy)/Z=lnx
So, Z / & y = x ^ y * LNX
So, &; Z / &; Y (1 / x, 1) = (- LNX) / X
әz/әy=-(csc(2xy^2))^2*4xy