Several problems with the limits of higher mathematics, problems with the limits of higher mathemati

The first question is itself a definition of e, and the proof of the limit convergence can be referred to the pee.

The second question is directly made: (n2+1) ln(1+ 2 n) ln(1+3 n)=[(n 2+1 ) (n 2) (n 3)]*ln[(1+2 n) (n 2)]*ln[(1+3 n) (n 3)]It is easy to know that the result is 6 (the last two limits are 1) The third question The limit should be 2 3 You estimate that the question is written incorrectly Divide the top and bottom by n to get 2 (3+1 n) The limit is obviously 2 3 Question 4 |(sin n）/n-0|<=|1/n|=1 n Any... Eight strands) take n=1.

Question 5 un The limit when n takes the positive infinity is a defined and given to >0 there exists n, arbitrary n>n, |un-a|< then there is a |︱un︱-|a||<=|un-a|< un The limit when n takes the positive infinity is a counterexample: un=(-1) n then the un limit is 1 , and the un limit obviously does not exist.

1) Newton's binomial theorem gives the expression of e, which is 1/1 of the reciprocal of the factorial from 0 to positive infinity. You look for advanced math or mathematical analysis, it's all there. This is to prove that the sequence is bounded and convergent.

2) In the second question, n tends to be positive infinity, n 2+1 can be replaced by n 2, and an infinite amount of bounded quantities is added to absorb the bounded quantities. Then n*n is assigned to each ln, and when you mention the exponent in ln, you will find that it is very similar to e, but the internal numerators 2 and 3, you can do the deformation. (3) n tends to positive infinity, 1 is absorbed and rounded, answer 2 3(4) sin n is a bounded quantity, 1 n is infinitesimal when n tends to positive infinity, infinitesimal multiplied by bounded quantity or infinitesimal, infinitesimal limit 0.

5) The sequence is positive and negative, such as -1, 1, -1, 1, -1, 1, ......, the limit does not exist, but the limit of the absolute value is 1. These are all basic calculus comparisons, and it is recommended to refer to the limits section of the calculus textbook.

The important limit lim(1+m) 1 m when m 0 is e lim(1+1 n) n when n 1 n 0 so the limit is e

Can be replaced: sin3x 3x tan5x 5x So, ruined stupidity and the limit is -3 5 and tell you about the reason why it can be replaced: We know sinx x (x 0) sin3x, set 3(x- )t, trace because x, so, file chang t 0

And 3x-3 = t, we get 3x=t+3 so sin3x=sin(t+3 )=sint -t

From the function image of y=sinx, we can see that when 0 x, y=sinx is positive, and when x 2, y=sinx is negative, so the left limit of y=sinx is 0+, and the right limit is 0-

You draw an image of y=sinx near , when x tends to be from the left, y tends to 0 from positive, and y tends to 0 from negative when x tends to from the right.

Answer question 1: Look at whether the number of the numerator is greater than 0 or less than 0, if the number of the numerator is greater than 0, there is "the left limit is negative infinity, and the right limit is positive infinity", then x=0 is the discontinuity point of the second type of infinity. If the number of the numerator is less than 0, then there is "the left limit is positive infinity and the right limit is negative infinity", then x=0 is still the discontinuity point of the second type of infinity.

In short, x=0 is the discontinuity point of the second type of infinity.

Answer question 2: The limit sinx x=1 cannot be removed and used directly in the operation.

1. Infinite discontinuity.

2. No, you cannot. sin x=1, only in the case of the limit. Remove the limit symbol and the equation does not hold.

Hello, what kind of math problem is it?,As long as it's not ridiculously difficult, I can try it.,Oh, it's not graduate math.。

If it's a graduate school entrance examination, you can.

Can you send out the question and take a look?

Question: Why can't this numerator replace (sinx) 2 with the equivalent infinitesimal with x 2 Although the above is subtraction, the numerator is replaced by a fourth denominator, and the accuracy is enough.

Question, but his numerator and denominator are not the same.

You can't add or subtract.

Ask a question, you look at this diagram.

Here's what you want to say.

You can also use the Lobida rule directly.

Equivalent infinitesimal is not to be used in addition or subtraction.

But the Taylor formula can be used.

It's McLaughlin-esque.

y=x-[x] is.

a unbounded function b is a function with a period of one.

c monotonic function d even function.

No, it should be done after grabbing.

Solution: original formula =sin( 6-x)*sin3x cos3x=sin( 6-x)*sin3x sin( 2-3x)=sin( 6-x)*sin3x sin3*( 6-x).

sin3x→1

sin( 6-x sin3*( 6-x) 1 3 so original = 1 3

For the sake of simplicity, the limit symbol in the middle is omitted).

Use the commutation method to make x=1 t known.

2 If you change it to 4, it's 2 2 3

See figure. Ah, I ran downstairs after changing it.

It is enough to consider only the coefficients of the highest order, and the other terms can be ignored in the limit process.

The first result is root number 2

The result of the second question is (2 30)*(3 20).

1、√(2x+1)-3＝[2x+1-9]/[√(2x+1)+3]＝2(x-4)/[√(2x+1)+3]

x-2)-2 (x-4) [ (x-2)+ 2]primitive lim 2[ (x-2)+ 2] [ (2x+1)+3] 2[ 2+ 2] [3+3] (2 2) 3

2. The numerator and denominator are divided by x 50 to get 2 30 3 20

There is nothing difficult about this problem, it is the basics.

The main thing is that you have a problem with this processing.

To meet the |1/x+2|>m, it is necessary to make |1/x+2|The minimum value of m

In the field of x=0, |1/x+2|The minimum value is |1/x|-2

Type 0 0 cannot be calculated directly.