How to understand 9 n 2 3 9 n when proving lim n 3n 2 n 2 3 3 3 with the limit definition

To prove with a definition that lim 3n 2 (n 2-3)=3 takes a look at it first. 3n^2 / (n^2-3) -3 ||3n^2-3n^2+9 / (n^2-3) |9 / |n^2-3|

Look again. n 2-3 = n, i.e., n 2-n-3 = 0 solution, n = (1 13) 2<3

Now limit n>3, then n 2-3>n

Nature, 9 (n 2-3) < 9 n

Thus, 3n 2 (n 2-3) -3 |<9 n to arbitrary. >0, there is n=max>0, and when n > n, there is always | 3n^2 / (n^2-3) -3 |<

So according to the definition, lim 3n 2 (n 2-3) = 3, and what you wonder is why add a step "9 (n 2-3)<9 n"?

This step is actually to simplify the operation, because when proving the limit with a definition, we have to find n, and this n is related to , and in the process of finding n, we need to solve the inequality.

In other words, if you don't do proper amplification and simplify the inequality before, it will be complicated to solve the inequality later.

Of course, it's okay not to simplify, as long as you can find out the relationship between n and .

If you don't understand, please ask.

Summary. Send the original question to the teacher.

lim n 2n 3n 6 3n n 3 find the limit.

Send the original question to the teacher. The fourth.

Evaluated according to the highest order.

Want to be a process senior.

Because the highest power order is the highest.

How to calculate.

Is there a process. Use the Lopida Rule.

No, it won't. You wait.

I'll do the math for you.

Considering the integral of the function f(x)=3 x on 0,1, divide the unit interval into n segments with a length of 1 n. Then the integral of f(x) is approximately [3 (1 n)+3 (2 n)+/n

n tends to positive infinity, which is equivalent to infinite subdivision, which is the definition of integral.

The answer is 2 ln3

The numerator and denominator are divided by 3 (n+1) at the same time

Original = lim[(1 3)(-2 3) n+1 3] [(-2 3) (n+1)+1]=(0+1 3) (0+1)=1 3

Proof: |(3n-2)/(2n+1)-3/2|=|7/[2(2n+1)]|7/4n<ε

Then n>4 7, take n=[4 7], and the parentheses represent the whole number.

Then for any given >0, there is always a positive integer n=[4 7], and when n > n, such that |(3n-2)/(2n+1)-3/2|< holds, so lim(3n-2) (2n+1)=3 2

Why didn't you write the approach range of n?

1 If n—the shed lifts the erection (tends to) infinity:

lim （3^n + 2^n）/（3^(n+1)-2^(n+1)）lim（3^n /3^(n+1) +2^n/3^(n+1)）/3^(n+1) /3^(n+1) -2^(n+1)/3^(n+1)）

lim(1 3 + 1 3)*(2 3) n) (chain size 1 - 2 3) (n+1)).

n-> infinite.

So (2 3) n=0; (2/3)^(n+1)=0.

lim(1 3 + 0) (1 - 0) = 1 32 if n->0

2/3)^n=2/3 ; 2/3)^(n+1)=4/9lim（1/3 + 1/3)*(2/3)^n）/（1 - 2/3)^(n+1)）

lim(1 3 + 2 9) 1 - 4 9)lim (5 9) 5 9) = 1