3 mathematical analysis proof questions! (real analysis, sequence, limit) When n approaches infinity, the sequence Sn / N = non-zero constant C. It is proved that the sequence Sn diverges to infinity The real number equation f is uniformly continuous on R. if FN (x) = f (x + 1 / N), it is proved that the sequence FN is always continuous and tends to F Proof: there is no continuous equation G: R - > R, so that G (x) = C has exactly two solutions (C is any real number) The second question should be: verify that the sequence FN is uniformly continuous and tends to F

3 mathematical analysis proof questions! (real analysis, sequence, limit) When n approaches infinity, the sequence Sn / N = non-zero constant C. It is proved that the sequence Sn diverges to infinity The real number equation f is uniformly continuous on R. if FN (x) = f (x + 1 / N), it is proved that the sequence FN is always continuous and tends to F Proof: there is no continuous equation G: R - > R, so that G (x) = C has exactly two solutions (C is any real number) The second question should be: verify that the sequence FN is uniformly continuous and tends to F

1. Let C > 0, because LIMS [n] / N = C, there is n, when n > = n '| s [n] / n-c| CN / 2
Any given positive number m, as long as n = max {n ', 2m / C}, when n > = n, s [n] > CN / 2 > = m, so s [n] diverges to infinity
2. For any given positive number a, there is a positive number B, when | x-x '| so for any a, when | x-x' | = | (x + 1 / N) - (x '+ 1 / N) | because f (x) is continuous, take an n > 1 / B for any a, and when n > = n, | (x + 1 / N) - x| 3, assume existence
G (x) = 0 has two solutions, let it be a, B (a). If 0 is both the maximum and minimum, it is contradictory
If 0 is neither the maximum nor the minimum, then x1, X2 ∈ (a, b), there is a X3 between X1 and X2, so that G (x3) = 0, which is contradictory
Therefore, 0 is either the maximum value or the minimum value. It is advisable to set it as the minimum value. Because g (x) = m has exactly two solutions X1 and x4, there is either one solution or two solutions in [a, b]
If there is one solution, then X4 is outside [a, b]. Let X4 > b and take M '∈ (0, m), then G (x) = m' has solutions in (a, x1), (x1, b), (B, x4)
If you have two solutions, you might as well set x1, so it doesn't exist

Prove LIM (n ^ 2 + N + 1) / (2n ^ 2 + 1) = 1 / 2 by sequence limit

For any ε> 0
Let n = max (1,3 / (4 ε))
When n > n
|(n^2+n+1)/(2n^2+1)-1/2|
=|2n^2+2n+2-2n^2-1|/[2(2n^2+1)]
=(2n+1)/[2(2n^2+1)]
Molecule 2n + 12 (2n ^ 2) = 4N ^ 2

Find the limit of the following sequence That's it ==

These topics are far from representative. The reason why you can't do your homework at ordinary times may be that you know the basic things. In fact, there are some limits of a function I use to solve the problem below in the book. Remember these clearly, and know how some basic forms change. There is no problem in general seeking the limit! The following is the process of solving these problems. I have written for a long time. I hope you can summarize and study them yourself,
1： It is known that the formula satisfies the conditions for using the norbida rule. Therefore, find the first-order reciprocal of the numerator denominator of function f to obtain 4x ^ 3 / (3x ^ 2) = 4x / 3. Since x tends to 1, the limit is 4 / 3
2： Multiply the numerator and denominator of the integrand function by √ (x) at the same time+ Δ x) + √ (x) substituted after simplification Δ X = 0, the limit is 1 / (2 √ x)
3： This problem is the same as the second problem. First, multiply the numerator and denominator of the function by √ (2x + 1) + 3, and then substitute x = 4 to get the limit as infinity
4： This problem is the same as the two and three solutions. After simplifying the numerator and denominator of the function by 1 + √ (TaNx + 1), substitute x = 0 to get the limit of - 2
6： The numerator denominator (here, the denominator is regarded as 1) is multiplied by √ (x ^ 2 + X + 1) + √ (x ^ 2-x + 1) to simplify the result. After simplification, 2x / (√ (x ^ 2 + X + 1) + √ (x ^ 2-x + 1)) is obtained. When the numerator denominator of the formula is divided by X, the numerator becomes 2 and the denominator becomes √ (1 + 1 / x + 1 / x ^ 2) + √ (1 - 1 / x + 1 / x ^ 2). Because x is infinite, the formula √ (1 + 1 / x + 1 / x ^ 2) + √ (1 - 1 / x + 1 / x ^ 2) All terms containing x in tend to 0. At this time, we can directly regard it as 0, and get that √ (1 + 1 / x + 1 / x ^ 2) + √ (1 - 1 / x + 1 / x ^ 2) tends to 2, so the whole limit is 1 / 2
7： Because the SiNx function is bounded, when x tends to 0, we can use the conclusion that the value of a bounded function is still infinitesimal multiplied by infinitesimal, that is, the limit of this problem is 0
8： This topic first multiplies the denominator by 2 (and then multiplies the whole function by 0.5 to ensure that the function value is the same as the original). The obtained form is the standard form above the advanced mathematics textbook: when x tends to 0, the limit of SiNx / X is 1, similar, so the limit is 0.5
9： When x tends to 0, 1-cosx can be regarded as the high-order infinitesimal of x ^ 2 / 2, and TaNx is the high-order infinitesimal of X, that is, the numerator becomes x ^ 2 / 2 and the denominator becomes x ^ 2, so the limit is 1 / 2
10： Because when x tends to 0, 2x also tends to 0, tan2x can be regarded as 2x, then the denominator of the function can be written as 2x, and then multiply the numerator denominator of the transformed function by √ (1 + x) + 1 to obtain the limit of 1 / 2
11： (2x + 1) / (2x-1) is reduced to 1 + 1 / (x-0.5), so that t = x-0.5 (x tends to infinity, so t also tends to infinity), that is, x = t + 0.5, so the original function can be written as the (T + 0.5) power of (1 + 1 / T), which can be written as: (1 + 1 / T) ^ t multiplied by (1 + 1 / T) ^ 0.5. When t tends to be infinite, the limit of (1 + 1 / T) ^ t is e, (1 + 1 / T) ^The limit of 0.5 is 1. Multiply the two limits to obtain the required limit, and the result is e
12： It is known that when x tends to infinity, the limit of the X-Power of (1 + 1 / x) is e (in the textbook), and the inherent X-Power of (1-1 / x). When x tends to infinity, the limit is 1 / E (all this type of problems use this solution). Moreover, the limit of the original function can be regarded as the k-power of the limit of the X-Power of (1-1 / x) (one of the properties of the limit), so the limit is 1 / e ^ K

A few proof questions about the limit of sequence, do me a favor... Lim will omit and don't type n/(n^2+1)=0 √(n^2+4)/n=1 sin(1/n)=0

The essence is the calculation problem, but the problem tells you the answer. Just write the process. The first problem, divide the numerator and denominator by N, and bring N equal to infinity into it. The second problem uses Heine's theorem to replace n with X. the original problem changes from sequence limit to function limit, and uses Robida's Law (I don't know whether the landlord has learned it or not, if he hasn't learned it, then

Find the limit of the sequence, LIM (n tends to infinity) (3N + 5) / n square under root sign + N + 4 =?

LIM (n tends to infinity) (3N + 5) / n square + N + 4 under the root sign=
Both numerator and denominator are divided by n
namely
LIM (n tends to infinity) (3 + 5 / N) / root sign (1 + 1 / N + 4 / N) ²）= 3/1=3
among
When LIM (n tends to infinity), 5 / N = 0 1 / N = 0 4 / n ²= 0

1. A team is 120m long. When the team is in progress, the correspondent rushes to the front of the team from the end of the team, but returns to the end of the team immediately. If the team advances 288M and the speed of the team and the correspondent remains the same during this period, what is the distance the correspondent walks during this period? 2. The propagation speed of sound in the air is 340m / s, and the propagation speed in steel is 5200m / s. Zhang Bo and Li Feng designed a scheme to measure the length of railway tunnel. Zhang Bo took the stop watch to the other end of the tunnel. Li Feng knocked on the steel rail with a hammer. Zhang Bo recorded that the time interval when he heard the knocking sound was 5.4s. Please help them find out the length of the tunnel

1. If you take the team as the reference system, you can easily get: the correspondent has walked 240 meters, but now the reference system has moved 288 meters, so the distance of the correspondent is 240 + 288 = 528 meters. If you don't understand, you can think so. You have walked 240 meters on the earth, and the earth has walked 288 meters, so... 2