To prove the limit of a sequence, it is necessary to prove it with a monotonous definite theorem

First, it is proved that there is an upper bound, that is, for any n, xn is less than or equal to some constant c.

We prove that xn<=2 is mathematically inductive.

2;2.Let xk<=2,x(k+1)= (2+x(k))<= (2+2)=2;

xn<2;

Then prove that xn is monotonically increasing:

If we already know that xn<=2, then xn= (2+x(n-1))>= (x(n-1)+x(n-1))) = 2*x(n-1)>=

x(n-1)*x(n-1)=x(n-1)；The above derivations are based on x(n-1)<=2

So xn>=x(n-1), so xn is a monotonic increment sequence.

The above proves that the XN sequence monotonically increases and has an upper bound, so the limit exists.

In fact, the limit of this series is 2, and the calculation limit can be calculated in this way.

Let x be the limit of xn, and take the limit on both sides of the equation xn = (2+x(n-1)).

x= (2+x), and the solution is x=2, and x=2 can be known

an>0 from the title

Take bn=1 an, then bn+1= [(an+1) an]= (1+1 an)= (1+bn).

As the characteristic equation x= (1+x), x>0, the solution is x=(1+ 5) 2

bn+1-(1+√5)/2|=|√(1+bn)-(1+√5)/2|

1+bn-(1+√5)²/4|/|√(1+bn)+(1+√5)/2|

bn-(1+√5)/2|/|√(1+bn)+(1+√5)/2|

bn-(1+√5)/2|/|(1+√5)/2|

5-1)/2*|bn-(1+√5)/2|

0<|bn-(√5+1)/2|<(5-1)/2*|bn-1-(1+√5)/2|<.5-1)/2]^(n-1)*|b1-(1+√5)/2|

lim(n)5-1) 2] (n-1)*|b1-(1+√5)/2|=0

lim(n→∞)0=0

The entrapment theorem yields lim(n)bn-( 5+1) 2|=0

lim(n→∞)bn=(√5+1)/2=1/an

lim(n→∞)an=2/(√5+1)=(√5-1)/2

Because the function is bounded, the range of values of the function is bounded.

So there must be a "minimum upper bound" (supreme) in the range of functions, and because s is a monotonic function, it corresponds to an arbitrarily small e>0, and there must be n>0 such that for any x>n, there is | f(x) -s | e

Meet the definition of the limit.

Hope it helps.

(o ) Good luck in your studies

The explanation of this theorem in the Tongji textbook is: We do not prove this theorem, but only give its geometric meaning on the number line, you can refer to it. If you want to test this question, you will not test the theorem proof, but you will first prove the monotonicity of a certain number series, and then prove the boundedness of this number series, so as to conclude that this number series must be convergent, that is, there is a limit, and then take the limit assumption a on both sides of the known equation that the number series satisfies, and then solve the equation a, which is the limit value of the number series.

To put it simply, it is to follow this criterion and then find two conditions to explain the existence of the limit, and then calculate the limit value.

The following introduces the same principle for single increase and single decrease.

1. The monotonous and bounded introduction limit of the series exists.

2. The limit existence cannot push the monotonic bounded number series, such as (-1) n*1 n.

3. Sufficient and unnecessary conditions.

Each item in the exponential column of the bounded number series does not exceed a fixed interval with an upper bound and a lower bound.

Suppose there is a fixed value a, and any n has a lower bound b, and if both a and b exist so that the value of the series is in the interval [a,b]bai, the series is bounded.

If the sequence satisfies: for all n there is xn m (where m is a constant independent of n) and says that the sequence is bounded (there is an upper bound) and that m is one of his upper bounds.

Sufficient but not necessary.

First, it is proved that there is an upper bound, that is, for any n, xn is less than or equal to some constant c.

We prove that xn<=2 is mathematically inductive.

2;2.Let xk<=2,x(k+1)= (2+x(k))<= (2+2)=2;

xn<2;

Then prove that xn is monotonically increasing:

If we already know that xn<=2, then xn= (2+x(n-1))>= (x(n-1)+x(n-1))) = 2*x(n-1)>=

x(n-1)*x(n-1)=x(n-1)；The above derivations are based on x(n-1)<=2

Therefore, xn>=x(n-1), so xn is a monotonic increase sequence, the above proves that the xn sequence has an upper bound for monotonic increase, so the limit exists, in fact, the limit of this series is 2, and the calculation limit can be calculated in this way Let x be the limit of xn, and take the limit on both sides of the equation xn= (2+x(n-1)) to have x= (2+x), and solve x=2, we can know x=2

an>0 from the title

Take bn=1 an, then bn+1= [(an+1) an]= (1+1 an)= (1+bn).

As the characteristic equation x= (1+x), x>0, the solution is x=(1+ 5) 2bn+1-(1+√5)/2|=|1+bn)-(1+√5)/2|=|1+bn-(1+√5)²/4|/|1+bn)+(1+√5)/2|=|bn-(1+√5)/2|/|1+bn)+(1+√5)/2|<|bn-(1+√5)/2|/|1+√5)/2|=(√5-1)/2*|bn-(1+√5)/2|0<|bn-(√5+1)/2|<(5-1)/2*|bn-1-(1+√5)/2|<.5-1)/2]^(n-1)*|b1-(1+√5)/2|

lim(n)5-1) 2] (n-1)*|b1-(1+√5)/2|=0

lim(n→∞)0=0

The entrapment theorem yields lim(n)bn-( 5+1) 2|=0∴lim(n→∞)bn=(√5+1)/2=1/an∴lim(n→∞)an=2/(√5+1)=(5-1)/2

Proof: Because there is an upper bound in the number series, then there is an upper bound a, and for any or socks > 0, a- is not an upper bound, so it is good to tease in n, so that a-

1. If it is proved that the sequence is monotonically increasing and has an upper bound, then the sequence has a limit.

2. If it is proved that the sequence is monotonically decreasing and has a lower bound, then the sequence has a limit.

Let the monotony be bounded (you may wish to set a single increase), then there is m>=x[n] (any n), so there is an upper definite boundary, which is denoted as l

For any positive number a, there is a natural number n, such that x[n]> l-a because x[n] is single-incremented, so when n>=n, l-a

This is not only monotonous, bounded, but also a conditional proof.

But this question is monotonically bounded except for the first one, so it can be proved with monotonous boundedness.

x1=√(2+a)《2

x(n+1)= 2+xn)《 2+2)=2 xn has an upper bound 2

x2=√(2+x1)=√2+√(2+a))》2+a)=x1

x(n+1)= 2+xn)" 2+xn-1)=xn xn single increase.

2x1=√(2+a)>2

x(n+1)= 2+xn)> 2+2)=2 xn has a lower bound 2

x2= (2+x1)= 2+ (2+a)),1,Are you sure you didn't make a mistake in the question,0,When 0 When a=2, it's always 2.Limits exist.

When a>2, the monotonic decreases, but xn>=2Monotony is bounded, so the limit exists.

The limits are all 2The following is a prayer:

According to xn+1=(2+xn), xn+1 2=2+xn, when n tends to infinity, because the limit of stupid grinding exists, so xn+1=xn

So it becomes x 2-x-2 = .0, Advanced Mathematics - Uses monotonic bounded criterion to prove that the limit of the number series exists.

Let a>0, x1 = root number (2+a), xn + 1 = root number (2 + xm) Prove: lim n-> infinity xn exists, and find its value.