A postgraduate examination seeks the ultimate basic questions for teaching

Landlord, are you going to graduate school? I must be able to do this kind of problem at this stage. This is a simple problem that uses Taylor series to find the limit (the postgraduate examination question is only more difficult than this).

First of all, the denominator is replaced by the equivalent infinitesimal to become x 3, and based on this, we only need to put the numerator of the general division to the 3rd order of x's Taylor formula (it has no practical meaning at the higher order, [because the denominator is only 3rd order]). There is a 0 (x 4) in the middle because it is replaced with a variable. Since the textbook only gives the formula e x, you can replace x with x 2 in this problem, and then simplify it with an infinitesimal computational relation.

Good luck!

The first step of the general fraction, equivalent infinitesimal substitution, should be fine, this part is used to deal with the denominator.

The key is the second step, the processing of the numerator, this problem uses the method of series, the denominator is x 3, so as long as the order is greater than 3, that is, the infinitesimal higher order than x 3, you can write it as o(x 4), in the operation, you can directly simplify it to 0.

You still have to lay a solid foundation, take a hard look at what weaknesses you have in computing, and do the questions you can do correctly, and you should be able to get the test to about 135.

I wish you success in your graduate school entrance examination.

Now that you're going to graduate school, you still don't know what McLaughlin's formula is? This is the basic content of the number of points, and it is introduced in any number book.

This kind of problem belongs to the factorization type of solving the limit problem, just pay attention to the skills and methods.

Because the limit exists, the molecule (x 2-3x+k) = 0, when x = 5;k = -10

Answer: The Ming Dan case is 0, and the auspicious disturbance process is as follows, please refer to it.

lim ( n-> infinity)[2 (n+1)+3 (n+1)] [2 n+3 n].

lim ( n-> infinity)2 (n+1) [2 n+3 n] +lim ( n-> infinity)3 (n+1) [2 n+3 n].

lim ( n-> infinity) (2 [ 1+ (3 2) n] +lim ( n-> infinity) 3 [ (2 3) n +1] )

Mainly there are physical and chemical factors, the original numerator and denominator are multiplied by (3x 3+4x+1)+(ax+b), the limit is 0, then the numerator is constant, then (ax+b) 2 3x 2+4x+b 2, obviously a 2=2,2ab=4,a= 3,b=2 3=2 3 3,

The method is shown in the figure below, please check it carefully, and I wish you a happy study: