Very simple algorithm, because there is no determinant here, so you have to search online yourself, and the algorithm is easy to remember.
Limit: Seek the limit according to L'H?pital's law (seek the amorphous limit of 0/0 type and ∞ /∞ type).
lim f(x)/g(x)=lim f'(x)/g'(x)
For example, x→0, lim sinx/x=lim cosx= 1, of course, it is not that difficult.
Generally x→2, lim (x2-3x+2)/(x-2) = lim (2x-3) =1.
Function: the derivative rule of implicit function, that is, the derivative rule of composite function.
Xy= 1, the derivatives of both sides y+xy'=0, y' =-y/x =- 1/x 2.
Series (series parts):
1. The limit of the ratio of the latter item to the former item is the proportional common ratio (see D'Alembert method for details).
For example, to prove Sn
q = lim a & ltn+ 1 & gt； /a & lt； N>, q< 1, and then a.
A<n>& ltBmq (n-m), (converted according to item m)
2. Find the limit of recursive sequence with fixed points (mainly used to discuss the exact range)
The most common is
If you can prove that one
3. Solution of homogeneous linear recurrence formula (difference equation)
This method is very fast, but it can't be used for calculating problems in high school. Verifiable.
Generally speaking, a< is the most, and it is a second-order A.
Construct the equation x 2+px+q = 0.
1. Two x 1, x2, and then one < n >；; General solution a < n > = c1(x1) n+C2 (x2) n.
(Note that x 1 and x2 can be complex numbers. )
2. Re-root x0, and then
C 1 and C2 are undetermined coefficients, and the known values of these two terms are substituted in the general solution, generally a.
take for example
Example 1:
a & ltn+2 & gt； -a & lt； n+ 1 & gt； -a & lt； n & gt=0，a & lt 1 & gt； = a<2> (Fibonacci series)
X 2-X- 1 = 0, and the solution is x 1=( 1+√5)/2, and x2=( 1-√5)/2.
So a < n >；; =c 1[( 1+√5)/2]^n+c2[( 1-√5)/2]^n
replace
a & lt 1 & gt； = 1 = c 1( 1+√5)/2+C2( 1-√5)/2
a & lt2 & gt= 1=c 1[( 1+√5)/2]^2+c2[( 1-√5)/2]^2
That is, C 1 and C2 are all solved.
So as to obtain a
Example 2:
a & ltn+2 & gt； -4a & lt； n+ 1 & gt； +4a & lt； n & gt=0，a & lt 1 & gt； =2，a & lt2 & gt=4
X 2-4x+4 = 0, and multiple roots x0=2.
General solution a < n > = (c1+C2 * n) 2 n.
a & lt 1 & gt； =2=(C 1+C 1)2
a & lt2 & gt=4=(C 1+2C2)2^2
Solve C 1, C2, and get a < n >；;
Inequality: Cauchy inequality (rarely involved) has many forms.
That's about it. Other methods are not easy to operate, some are not competition knowledge, but some basic knowledge of college mathematics.
These methods must indicate the source (theorem name, etc.). ) take the exam, or you will lose points.