Ask: how to calculate, when x approaches 0, 1 / X * sin (1 / x) approaches infinity?

Ask: how to calculate, when x approaches 0, 1 / X * sin (1 / x) approaches infinity?

Let u = 1 / x, then when x → 0, u →∞
So Lim 1 / X * sin1 / x = Lim usinu
Let u be 2 π, 4 π, 6 π,..., 2K π,..., respectively,
The subcolumn of function value is 0,0,..., 0... Tends to 0
Let u be π / 2,..., 2K π + π / 2
The subsequences of function values are, π / 2,..., 2K π + π / 2,..., tending to ∞
The convergence of two subsequences is different
So this limit doesn't converge, but it's not infinite

When X - > 0, the variable (1 / x ^ 2) sin (1 / x) is unbounded, but not infinite. Why?

Roughly speaking: X - > 0, then 1 / X tends to infinity, sin (T) is a periodic function, so no matter how much T increases (decreases), there is always a change of - 1 - + 1 near it, so it does not converge, plus the (1 / x ^ 2) part, it becomes unbounded [because (0 + EPS) multiplied by infinity, no one is sure how much, EPS represents a tiny positive number]

Given that {an} is an increasing sequence and an = n ^ 2 + λ n holds for any n ∈ n *, then the value range of real number λ is
Why don't you count - in / 2 < 1?
But the answer is more than 3
I'm here to thank you
The sequence is discrete and the function is continuous, but it is not possible to find the maximum value

an+1=(n+1)^2+λ(n+1)
an+1-an=2n+1+λ
If it is an increasing sequence: 2n + 1 + λ > 0
λ> - (2n + 1) constant set up
λ>-3
The answer is wrong: for example, when λ = 0, an = n ^ 2 is also increasing
The sequence is discrete and the function is continuous, so the method of quadratic function cannot be applied to the sequence
Add another question: No, for example, an = - n ^ 2 + N, the symmetry axis of quadratic function is 1 / 2, but we can't get this point on integers

If {an} is known to be an increasing sequence and an = N2 + λ n is constant for any n ∈ n *, then the value range of real number λ is ()
A. （-72，+∞）B. （0，+∞）C. [-2，+∞）D. （-3，+∞）

∵ {an} is an increasing sequence, ∵ an + 1 ＞ an, ∵ an = N2 + λ n is constant into immediate (n + 1) 2 + λ (n + 1) ＞ N2 + λ n, ∵ λ ＞ - 2N-1 holds for n ∈ n * constant. When n = 1, the maximum value of - 2N-1 is - 3, ∵ λ ＞ - 3, so D is selected

Given that the sequence {an} is an increasing sequence, and for any positive integer n, an = n & # 178; - λ n is constant, then the value range of real number λ is?

A(n+1)-A(n)
=2n+1 -λ >0
λ< 2n+1
λ < 3

It is known that {an} is an increasing sequence, and for any n belonging to n *, an = n ^ 2 + λ n is constant, then the value range of real number λ is?
This is 2010 Yangzhou simulation, emergency!

I have answered you once
A(n+1)-An=(n+1)^2+λ(n+1)-n^2-λn (n=1,2,.)=2n+1+λ
We can know that when n = 1, a (n + 1) - an is the smallest, and the sequence is an increasing sequence, so a (n + 1) - an > 0
That is, a (n + 1) - an > a2-a1 = 3 + λ > 0
λ>-3

Given that {an} is an increasing sequence and an = n ^ 2 + λ n holds for any n ∈ n *, then the value range of real number λ is
an+1=(n+1)^2+λ(n+1)
an+1-an=2n+1+λ
If it is an increasing sequence: 2n + 1 + λ > 0
λ> - (2n + 1) constant set up
λ>-3
How can we know λ > - 3 after finding λ > - (2n + 1) in the answer,

n∈N*
n=1、2、3、4..n
Do you understand natural numbers
When n = 1, λ is the maximum

For X ∈ R, the inequality | 2-x | + | 1 + X | ≥ a2-2a holds, then the value range of real number a is______ ．

∵ for X ∈ R, the inequality | 2-x | + | 1 + X | ≥ a2-2a holds, and the minimum value of | 2-x | + | 1 + X | is greater than or equal to a2-2a. Since | 2-x | + | 1 + X | represents the sum of the distances from the corresponding points of x on the number axis to the corresponding points of 2 and - 1, its minimum value is 3, so there is 3 ≥ a2-2a, that is a2-2a-3 ≤ 0, the solution is - 1 ≤ a ≤ 3, so the value range of real number a is − 1, 3, so the answer is − 1, 3

If a = {x | - 3 ≤ x ≤ 4} B = {x | 2m-1 ≤ x ≤ m + 1, m ≤ 2} a ∩ B = B, find the value range of real number M

Because m ≤ 2, 2m-1 ≤ m + 1, B is not an empty set
Because a ∩ B = B
Then - 3

If the line y = K (X-2) + 4 and the curve y = 1 + have and only have one common point, the value range of the real number k is obtained
If the line y = K (X-2) + 4 has and only has one common point with 4 - (x-square) under the root of the curve y = 1 +, the value range of the real number k is obtained

If the line y = K (X-2) + 4 and the curve y = 1 + √ (4-x ^ 2) have and only have one common point, the curve y = 1 + √ (4-x ^ 2). Y-1 = √ (4-x ^ 2). (Y-1) ^ 2 = 4-x ^ 2 (Y-1) ^ 2 + x ^ 2 = 4 is a circle, the center of the circle is (0,1), the radius is 2, so y = 1 + √ (4-x ^ 2) is a semicircle, and the part of the line above y = 1 crosses the point (2,4)