When x tends to positive infinity, Lim ln (1 + e to the x power) is divided by the root 4 times the square of X + 1

When x tends to positive infinity, Lim ln (1 + e to the x power) is divided by the root 4 times the square of X + 1

X tends to be positive infinity
One
ln(1+e^x)≈ln(e^x)=x
2.√(4x^2+1)≈√(4x^2)=2x
When x tends to positive infinity
limln(1+e^x)/√(4x^2+1)=x/2x=1/2
You can also use the law of lophida

When x tends to infinity, LIM (1-2 / 2x + 1) is to the power of X

lim(x-->∞) [1 - 2/(2x + 1)]^x
= lim(x-->∞) [1 - 2/(2x + 1)]^[- (2x + 1)/2 · - 2x/(2x + 1)]
= e^lim(x-->∞) - 2x/(2x + 1)
= e^lim(x-->∞) - 2/(2 + 1/x)
= e^(- 1)
= 1/e

Using LIM (1 + 1 / N) nth power X - > 0 = e to find the limit LIM (1-1 / N) nth power X - > infinity

The limit process should be n →∞, based on which:
lim(1-1/n)^n=lim[1+1/(-n)]^[(-n)·(-1)]
=lim[1+1/(-n)]^(-n)]^(-1)
=e^(-1)=1/e

LIM (n →∞) [2 (n + 1) power + 3 (n + 1) power] / [2n power + 3N power] limit Lim [2 (n + 1) power + 3 (n + 1) power] / [2n power + 3N power] limit (n →∞)

I'm just passing by to see your question. I don't know if it's just for reference
The original formula 2 ^ (n + 1) + 3 ^ (n + 1) / (2 ^ n + 3 ^ n) is not true
=2 ^ (n + 1) / (2 ^ n + 3 ^ n) + 3 ^ (n + 1) / (2 ^ n + 3 ^ n)
=2/(1+(3/2)^n))+3/((2/3)^n+1)
When n approaches infinity, 2 / (1 + (3 / 2) ^ n)) approaches 0 3 / ((2 / 3) ^ n + 1)
Approach 3
So the limit is three

Using the definition of sequence limit, it is proved that LIM (n - > ∞) 1 / (the k-th power of n) = 0

Xn=1/n^k
|Xn-a|=|1/n^k-0|=1/n^k<1/n
For any given positive integer ε (let ε < 1), as long as
1/n<ε,n>1/ε,
Then the inequality | xn-a | ε must be true. Therefore, if n > N, the positive integer n = [1 / ε] will exist
|1/n^k-0|<ε
There are:
lim（n->∞）1/n^k=0

Find the limit of the following sequence: LIM (n →∞) 2 to the nth power + 3 to the (n + 1) power / (divided by) 2 to the N + 1 power - 3 to the n power Find the limit of the following sequence: LIM (n →∞) 2 to the nth power + 3 to the (n + 1) power / (divided by) 2 to the N + 1 power - 3 to the n power

Divide both the numerator and denominator to the nth power of 3
The molecule is (2 / 3) to the nth power plus 3
The denominator is twice (2 / 3) to the nth power minus 1
When n →∞, the nth power of (2 / 3) tends to 0
2 times (2 / 3) is also close to 0
So the answer is - 3
The process is not easy to write... Some symbols do not know how to express
You'll know how to write it by yourself

By using the exact definition of sequence limit, it is proved that when n →∞, LIM (the power of N / the power of 2) = 0 Using the exact definition of sequence limit, it is proved that when n →∞, LIM (the power of N / the power of 2) = 0 Using the "exact definition proof" of sequence limit,

It is proved that for any ε > 0, the inequality is solved
When N 2 / 2 ^ n │ = n? 2 / (1 + 1) ^ n = n? 2 / [1 + N + n (n-1) / 2 + n (n-1) (n-2) / 6 +.] n, there is │ n? 2 ^ n │∞) (n? 2 / 2 ^ n) = 0

Find the limit LIM (x → 0) sinxsin (1 / x); LIM (x →∞) (arctanx / x)

lim(x→0)sinxsin(1/x)=0
[infinitesimal SiNx times bounded function sin (1 / x)]
lim(x→∞)(arctanx/x)=0
[same reason, arctanx bounded, 1 / X infinitesimal]

Limit: LIM (x → + ∞) (2 / π arctanx) x

This problem is wrong, the original question should be: find the limit: LIM (x → + ∞) (2 / π arctanx) ^ x solution 1: original formula = e ^ {LIM (x - > + ∞) [x (LN (arctanx) + ln (2 / π))]} (using the continuity and logarithmic properties of elementary functions) = e ^ {LIM (x - > + ∞) [(LN (arctanx) + ln (2 / π)) / (1 / x)]} = e ^ {LIM (x

LIM (2 / π arctanx) ^ x x x →∞ Lim x ^ 2 e ^ (1 / x ^ 2) x → 0, the limit is obtained by means of Robida

lim(2/πarctanx)^x =lime^[xln(2/πarctanx)]=lime^{[ln(2/πarctanx)]/(1/x)}=lime^{[1/(2/πarctanx)*2/π*1/(1+x^2)]/(-1/x^2)}=lime^{-x^2/[(1+x^2)arctanx]}=-2/π.lim x^2 e^(1/x^2) =lim e^(1/x^2)/(1/x^2)=...