Limit x approaches infinity (√ n + 3 - √ n) √ n-1

Limit x approaches infinity (√ n + 3 - √ n) √ n-1

Who knows how to do the proof of this pinch theorem?
Let A1 ≥ 0 AK ≥ 0, prove: n √ A1N + Ann + akn=max(a1,a2… ak)
Let A1 ≥ 0 AK ≥ 0, it is proved that: (n power of a1 + n power of an +...) The n-th root of ＋ AK = max (A1, A2 ak)

Let am be the largest in (A1, A2,..., AK). Obviously: (n power of a1 + n power of an + We know that the limit of K ^ (1 / N) is equal to 1 when n tends to infinity, that is, Lim [K ^ (1 / N)] = 1

What is the pinch theorem

If the sequence {xn}, {yn} and {Zn} satisfy the following conditions:
（1）yn≤xn≤zn(n=1,2,3,……） ,
（2）lim n→∞ yn =a,lim n→∞ zn =a,
Then the limit of sequence {xn} exists, and lim n →∞ xn = a

Prove the formula with the pinch theorem
How to prove that LIM (SiNx / x) = 1 when x tends to 0

F (x) = SiNx / X is an even function, so we only need to consider the right limit
Without proof, I hope the landlord will admit that:
0

What is the pinch theorem?

If the sequence {xn}, {yn} and {Zn} satisfy the following conditions: (1) from a certain term, that is, when n > n., where n. ∈ n, there is yn ≤ xn ≤ Zn (n = 1,2,3,...) (2) if n →∞, limyn = a; if n →∞, limzn = a, then the limit of sequence {xn} exists, and if n →∞, limxn = A. 2. If f (x) and G (x) are continuous in XO and have the same limit a, limf (x) = limg (x) = a, then if f (x) has f (x) ≤ f (x) ≤ g (x) in a neighborhood of XO, then if x approaches XO, If limf (x) ≤ limf (x) ≤ limg (x), that is, a ≤ limf (x) ≤ a, so limf (XO) = a simply: function a > b, function b > C, the limit of function a is x, the limit of function C is also x, then the limit of function B must be X. This is the content of Higher Mathematics: [application of pinch theorem in sequence] 1, The {Zn} is a convergent sequence, and the limit of {xn} and {Zn} is: A. if there is n, so that when n > N, there is xn ≤ yn ≤ Zn, then the sequence {yn} converges, and the limit is a. 2. The pinch criterion is suitable for solving the functional limit which can not be solved directly by limit algorithm, and the limit of F (x) can be determined indirectly by finding the limit of F (x) and G (x)

Lim nsin (3x / N) =? N tends to infinity

sin(3x/n)
lim----------- * lim(3x) =1*3x=3x
3x/n

Find LIM (n → + ∞) nsin π [√ (n ^ 2 + 2) - n]

N tends to infinity, find the limit of (1 + 2 ^ n) ^ (1 / N)

lim(n→∞) (1+2^n)^(1/n)
=lim(n→∞) e^{ln[(1+2^n)^(1/n)]}
=lim(n→∞) e^[ln(1+2^n)/n]
=e^ lim(n→∞) [ln(1+2^n)/n]
=e^ lim(n→∞) [(2^n)*ln2/(1+2^n)]
=e^(ln2)=2

n(1/1+n^2+1/4+n^2+1/9+n^2+… +1 / N ^ 2 + n ^ 2) when n approaches infinity, how to use definite integral to find limit
Attention is to use definite integral to find limit

Original form=
(n->∞)limn∑(i=1~n)1/(i^2+n^2)
=(n->∞)lim1/n∑(i=1~n)1/((i/n)^2+1)
=∫(0~1)1/(1+x^2)dx
=arctanx|(0,1)
=π/4

When n approaches infinity, find 1 / 1 × 2 + 1 / 2 × 3 + +The limit of 1 / N (n + 1)

1
1/(1×2)+1/(2×3)+...+1/[n(n+1)]=1-1/2+(1/2-1/3)+...+[1/n-1/(n+1)]
=1-1/(n+1)