The limit of the sequence 2 n 1 3 n

x+x^2+x^3+……You think of it as a proportional sequence, the common ratio is x, and then you use the equation to sum the proportional sequence.

The result of the limit calculation will have x n, and the problem will generally have a range of x values (0,1), so when n tends to infinity, the term containing x n will tend to 0.

The first limit is inside, the base.

And exponents are both variables about n, and in the second limit, the base is a constant x, and only the exponent n is a variable, so the limit is 0, and the two equations have nothing to do with each other.

Let x1 0 (approaching 0 is actually delta x).

y‘＝(x+x1）＾n-x^n)/x1

Then use the binomial.

Simplify the numerator to obtain: x (n-1) (x1) x (n-2)*(x1 2)+x (n-3)*(x1 3)+....x^0*(x1^n)

Divide by the denominator. Get x (n-1) x (n-2)*(x1)+x (n-2)*(x1 2)+....x^0*(x1^n-1)

Because x1 0

So y' (x+x1) n-x n) x1 x (n-1).

i.e. y' x (n-1).

2^n-1]/3^n = (2^n/3^n) -1/3^n = (2/3)^n - 1/3)^n.

n-> infinity, (2 3) n ->0, (1 3) n ->0, so, [2 n - 1] 3 n = (2 3) n - 1 3) n ->0-0 = 0

If the limits exist after separation, they can be found separately.

lim[(n-1)/(n+1)]^n

lim[(n+1-2)/(n+1)]^n=lim[1+(-2)/(n+1)]^n

lim[1+(-2)/(n+1)]^n+1-1)=lim[1+(-2)/(n+1)]^n+1)*1=^(-2)

According to the important limits: the correspondence of lim(1+1 n) n=e=e (-2)n.

Generally speaking, n gets bigger as it gets smaller, so n is often written as n( ) to emphasize n's dependence on change and change in . But this does not mean that n is uniquely determined: (e.g., if n> n makes |xn-a|< is true, then obviously n>n+1, n>2n, etc., also make |xn-a|< founded).

What matters is the existence of n, not the magnitude of its value.

When n=2n, [(1) n

1][(n+1)/n]

(-1)^(2n)

1]/[(2n+1)/(2n)]

2n+1)/n=2

1 n, n=2n-> infinity, n-> infinity, [(1) n+1][(n+1) n]=2

1 n>2n=2n+1, [(1) n

1][(n+1)/n]

(-1)^(2n+1)

1][(2n+2)/(2n+1)]=0.

n=2n+1-> infinity, [(1) n+1][(n+1) n]->

0 is not equal. 2.Thus, at n-> infinity, the limit of [(1) n+1][(n+1) n] does not exist.

lim[(n-1)/(n+1)]^n

lim[(n+1-2)/(n+1)]^n=lim[1+(-2)/(n+1)]^n

lim[1+(-2)/(n+1)]^n+1-1)=lim[1+(-2)/(n+1)]^n+1)*1=^(-2)

According to the important limit: lim(1+1 n) n=e=e (-2), the basic methods for finding the limit are:

1. In the fraction, the numerator and denominator are divided by the highest order, and the infinity is calculated by infinitesimal to infinitesimal and the infinitesimal is directly substituted by 0.

2. When the infinity root formula subtracts the infinite root formula, the molecule is rationalized.

3. Apply Lopida's law, but the application condition of Lopida's law is to become infinitely larger than infinite, or infinitesimal than infinitesimal, and the numerator denominator must also be a continuous derivative function.

When n=2n, [(1) n

1][(n+1)/n]

(-1)^(2n)

1]/[(2n+1)/(2n)]

2n+1)/n=2

1 n, n=2n-> infinity, n-> infinity, [(1) n+1][(n+1) n]=2

1 n>2n=2n+1, [(1) n

1][(n+1)/n]

(-1)^(2n+1)

1][(2n+2)/(2n+1)]=0.

n=2n+1-> infinity, [(1) n+1][(n+1) n]->

0 is not equal. 2.Thus, at n-> infinity, the limit of [(1) n+1][(n+1) n] does not exist.

Summary. Limit series lim [(n-1)(n+2)(2n-1)] 3n +n pro, hello, happy to serve you! I am Mr. Dong Xiaoming, and I am good at mathematics, physics and chemistry.

I will provide you with the process and answer the rough case within 5 minutes, please wait for a while.

Wait a minute, dear, I'll write down the calculation process on paper.

Look at the **, dear. The limit is 2 to 3.

The numerator and denominator can be divided by n at the same time.

Knowledge expansion: The test point of this question is to solve the limit grinding side state of the fraction. When encountering such problems, trying to divide the numerator and denominator by a blind source mononomial or polynomial at the same time often has obvious results.

-2 3) n-->0, so, (-2) n+3 n] permeate equilibrium nucleus[(-2) (n+1)+3 (n+1)]=2 3) n+1] [2(-2 3) n+3]--0+1] [0+3]=1 3

When n tends to be a positive infinite bush, the limiting limit of [(2) n+3 n] [2) (n+1)+3 (n+1)] is 1 3

n/(2n+1)=1/(2+1/n)

When n tends to infinity, 1 n tends to be 0, so the upper limit of the pole tung stall is 1 2

The number of beams is matched with 1 3, 2 5, 3 7 ,...n 2n plus 1 ,...The limit is 1 2n--> lim(1 n)=0, and n (2n+1)=(n n) (2n+1) n=1 (2+1 n), so n--> lim[n cavity slag (n+1)]=lim[1 (2+1 n)]=lim1 [lim2+lim(1 n)]=1 (2+0)=1 2

1/2 1/3 1/4 …Paiji....The limit of 1 n+1 is 0

3/2 4/3 5/4 ……The limit of n+2 n+1 is 1