Let X1 = 1, X2 = 1 + X1 / (1 + x1),..., xn = 1 + XN-1 / (1 + XN-1), find LIM (n tends to infinity) xn

Let X1 = 1, X2 = 1 + X1 / (1 + x1),..., xn = 1 + XN-1 / (1 + XN-1), find LIM (n tends to infinity) xn

x1>0→x2>0→…… →xn>0.
→0< xn-1/(1+xn-1)

It is known that the approximate value of a to the percentile is 5.28, and the approximate value of B to the thousandth is 6.246

The range of a and B is 245.65 and 245.65, respectively

It is known that the approximate value of a to the percentile is 5.28, and the approximate value of B to the thousandth is 6.246
It's a process

5.275

2.1678 to the approximation of the percentile
Such as the title

2.17

What is the limit x ^ (1 + x) / (1 + x) ^ x-x / E? (x approaches infinity)

The result is [(1 1 / x) ^ X-1 / E] / (1 / x), let t = 1 / X; get [(1 T) ^ (1 / T) - 1 / E] / T, t tends to 0, and then use lobita to get the result E

Kneel down: when x approaches infinity, find 1 / (n × n + N + 1) + 2 / (n × n + N + 2) + The limit of + n / (n × n + N + n)

Using the pinch theorem
Let s = 1 / (n × n + N + 1) + 2 / (n × n + N + 2) + ＋n／﹙n×n＋n＋n﹚
1／﹙n×n＋n＋n﹚＋2／﹙n×n＋n＋2﹚＋… ＋n／﹙n×n＋n＋n﹚≤S≤1／﹙n×n＋n＋1﹚＋2／﹙n×n＋n＋1﹚＋… ＋n／﹙n×n＋n＋1﹚
（1+2+...+n)／﹙n×n＋n＋n﹚≤S≤（1+2+...+n)／﹙n×n＋n＋1﹚
1/2*n(n+1)／﹙n×n＋n＋n﹚≤S≤1/2*n(n+1)／﹙n×n＋n＋1﹚
Using the pinch theorem to get the limit 1 / 2

When x = 3, the value of the algebraic formula ax to the 5th power + BX to the 3rd power + cx-10 is 7, then when x = - 3, the value of this algebraic formula can be obtained

When x = 3, the value of ax to the 5th power + BX to the 3rd power + cx-10 is 7
The fifth power of 3 is a + 3 & # 179; B + 3c-10 = 7
The fifth power of 3 A + 3 & # 179; B + 3C = 17
When x = - 3
Ax to the 5th power + BX to the 3rd power + cx-10
=-The fifth power of 3 A-3 & # 179; b-3c-10
=-(5 power of 3 A + 3 & # 179; B + 3C) - 10
=-17-10
=-27

Binary function limit problem!
lim ln(x+e^y)/sqrt(x^2 + y^2)
(x,y)->(1,0)

First of all, we can see that this limit must exist
In the case of existence, we can use the method of calculating the limit by stages
The original formula = LIM (x → 1) (Y → 0) ln (x + e ^ y) / sqrt (X & sup2; + Y & sup2;) = LIM (Y → 0) ln (1 + e ^ y) / sqrt (1 + Y & sup2;) = LN2

Find a limit of binary function,
Lim (x+y)[sin1/x]*[cos1/y]
(x,y)→(0,0)
x. Y tends to 0

Lim (x+y)[sin1/x]*[cos1/y] = 0
(x,y)→(0,0)
Because when x → 0, y → 0, x + y → 0 is infinitesimal,
And [sin1 / x] and [cos1 / y] are bounded,
According to the property of infinitesimal multiplied by bounded value or infinitesimal, the answer is 0

Limit of binary function
3-√9+xy
------------
xy
x. Y tends to zero
The dotted line in the middle temporarily replaces the fractional line

There should be no limit. If there is a limit, the limit should be the same in any direction. If we take x = y direction, the limit becomes a unary limit. Obviously, when XY - > 0, the denominator is 0 and the molecule is not 0, there must be no limit
If you don't add brackets after the root sign, no one can know that XY is in the root sign
If the root sign includes XY, then the above formula can be changed to
(3 + radical (9 + XY)) (3-radical (9 + XY)) / [XY (3 + radical (9 + XY))]
=- XY / [XY (3 + radical (9 + XY))] = - 1 / [(3 + radical (9 + XY))] = - 1 / 6