Math Limit Calculation Problems, Math Problems Limit Calculation Problems?

Is it n+3, or (n+3), if it is the former, the answer should be 5 to teach you a simple way, because n-> so 1 and 3 can be ignored, 2n 2=2, in fact, this is very natural, for this kind of simplicity you can see at a glance, of course, if you can't see it at once, divide the numerator by the denominator, and divide the polynomial by the polynomial, such as (2n-1) (n+3)=2-7 (n+3).

Take the limit to get 2

Keeping in mind the basic limits such as 1 n=0 (n->), these fundamentals are to be proved using the limit definition.

There's no need to get it that complicated, just imagine, if a number tends to infinity, then it doesn't make sense to be +1 or -2.

Considering the limit, you can remove the addition and subtraction process, and this problem directly regards it as 2n n=2 generations.

Conclusions from textbooks.

The limit of the fraction If the highest order of n is the same, then the limit value is the quotient of the coefficient of the highest order term, and if the highest order of the numerator is less than the denominator, then 0

If the highest order of the numerator is greater than the denominator, there is no limit value.

The expression (2n-1) (n+3) divides the numerator denominator by n and then finds the limit. The part where n is the denominator tends to be 0 when n tends to infinity

The detailed process macro changes are shown in the source figure RT.

I hope it is written very clearly.

Forehead ......Is there a mistake in this question? Hope it helps.

This belongs to the 0 0 limit problem, the numerator should be e 0-e 0=1-1=0, and one cannot be changed and the other is unchanged, so that it is wrong for the molecule to be e (x 2)-1 or 1-e (2-2cosx). The limit of the numerator and denominator of this problem is 0, so it is necessary to use the law to solve it, that is, the derivative method of the numerator and denominator respectively.

The principle of the same order up and down, the accuracy of cosx is not enough, the denominator is to the 4th power, and the numerator should also be to the 4th power

Molecule = [x (1 3)-1] 2

Denominator = (x-1) 2 = 2

After reduction, the original limit = 1 [x (2 3)+x (1 3)+1] 2 (x tends to 1)=1 3 2=1 9

The answer is 1 2021, done with the method of infinite series to integral.

Well, it's the right thing to do by making up definite integrals!

Note that there are positive minimum and maximum values for f(x), which may be denoted as m<=f(x).

Solution: limx 0 sin3x x=3x x=3, (equivalent infinitesimal )

limx 0 1 [ (9+sin3x)+3]=1 (3+3)=1 6, (directly bring x=0 in).

In summary, the original formula = 3 * 1 6 = 1 2

These are relatively simple limit problems, some of which are directly substituted, some use Lopida's rule, and some can be factored.

When x tends to infinity 1 x is an infinitesimal quantity, and arctanx is a bounded quantity, and the bounded quantity is multiplied by an infinitesimal quantity or an infinitesimal quantity, so the result is 0