A limit question: LIM (x approaching infinity) (2x+3/2x+1) x+1 power I know the answer is e do not know the specific process who can tell me next?

A limit question: LIM (x approaching infinity) (2x+3/2x+1) x+1 power I know the answer is e do not know the specific process who can tell me next?

Let t = (2x + 1) / 2, then let t = (2x + 1) / 2, then let t = (2x + 1) / 2, then let t = (2x + 1) / 2, then let t = (2x + 1) / 2, then let t = (2X + 1) / 2, then let t = (2x + 1 + 1) / 2, and x = T-1 / 2. Therefore, the original formula = LIM (t → ∞) (1 + 1 / T) (1 + 1 / T) ^ (T + 1 / 2) = Lim (t → ∞) (1 + 1 / T) ^ t * (1 + 1 / T) ^ (1 / 2) = e * (1 + 1 + 0) ^ (1 / 2) = e.. = = = = = = = = = = = = = = = = = = = = = = = =

Find the (x + 1) power of limit. LIM ((2x + 1) / 2x + 3)). X →∞

The original formula = LIM (x - > ∞) {[(2x + 3-2) / (2x + 3)] ^ (x + 1)}
=lim(x->∞){[(1+(-2)/(2x+3))^((2x+3)/(-2))]^[-2(x+1)/(2x+3)]}
=E ^ {LIM (x - > ∞) [- 2 (x + 1) / (2x + 3)]} (application important limit LIM (Z - > 0) [(1 + Z) ^ (1 / z)] = e)
=e^{lim(x->∞)[-2(1+1/x)/(2+3/x)]}
=e^[-2(1+0)/(2+0)]
=e^(-1)
=1/e.

LIM (n tends to infinity) [(a + N + C) / 3] n, a > 0, b > 0, C > 0

LIM (n - > ∞) {n * ln [(a ^ (1 / N) + B ^ (1 / N) + C ^ (1 / N)) / 3]} = LIM (n - > ∞) {ln [(a ^ (1 / N) + B ^ (1 / N) + C ^ (1 / N)) / 3] / (1 / N)} = LIM (x - > 0) {ln [(a ^ x + B ^ x + C ^ x) / 3] / X} (let x = 1 / N) = LIM (x (x - > 0) {[ln (a ^ x + X + C ^ x) / 3] / X} (make x = 1 / N) = LIM (x (x - > 0) {[ln (a ^ x + A ^ x + A ^ x + X + B ^ x + C ^ x) - Ln3] / X} = LIM (x - > 0) [(a ^ x * LNA + B ^ x * l

It is proved that the square root n (n tends to infinity) = 1

If n times root sign (n) = 1 + TN, then 0n = (1 + TN) ^ n = 1 + NTN + n (n -- 1) / 2 * TN ^ 2 +... > n (n -- 1) / 2 * TN ^ 2
TN ^ 2 < 2 / (n -- 1)
0 N times root sign (n) tends to 1

The nth power of LIM cosx / 2cosx / 4 cosx 2 (n tends to infinity)

Because cos X / 2cosx / 4 Cosx / 2 ^ n = [cosx / 2 * cosx / 4 *. * 2sinx / 2 ^ n * cosx / 2 ^ n] / (2sinx / 2 ^ n) = [cosx / 2 * cosx / 4 *... * SiNx / 2 ^ (n-1)] / (2sinx / 2 ^ n) = (cosx / 2sinx / 2) / [2 ^ (n-1) * sin (x / 2 ^ n] = SiNx / [2 ^ n * sin (x / 2 ^ n)], so LIM (n approaches positive infinity) C

What is the limit of the sum of the nth power of 2,3,4,5 and the root sign one nth of N

Take into account y = (2 ^ x + 3 ^ x + 4 ^ x + 5 ^ x) ^ (1 / x), then LNY = ln (2 ^ x + 3 ^ x + 4 ^ x + 5 ^ x) / x, when x tends to be positive infinity, y tends to be positive infinity, according to the law of Rhoda: ln (2 ^ x + 3 ^ x + 4 ^ x + 5 ^ x) / X tends to (2 ^ xln2 + 3 ^ xln3 + 4 ^ xln4 + 5 ^ xln5) / (2 ^ x + 3 ^ x + 4 ^ x + 5 ^ x) / (2 ^ x + 3 ^ x + 4 ^ x + 5 ^ x), the same division by 5 ^ x, the formula tends to LN5, the formula tends to LN5, the formula tends to LN5, the formula tends to LN5, the formula tends to LN5, so LNY tends to be LN5, y tends to be 5
So the limit of n tends to infinity is 5

N tends to infinity, the limit of (n + 1) power of ((2n + 3) / (2n + 1))

Teach you an important limit
For (1 + 1 / N) ^ n
Infinite time
(1 + 1 / N) ^ n = e ^ LIM (1 / N) * n, that is, LIM (1 + infinitesimal of n) ^ infinity of n = e ^ LIM (infinitesimal of n * infinity of n)
(n + 1) = e ^ LIM (2 / (2n + 1)) * (n + 1) = e ^ (1 / 2)

Just learn the limit of higher numbers, LIM (x → 1) [(x-1 to the nth power) divided by (x-1)]

The nth power of X-1 = (x-1) (the N-1 power of X + the N-2 power of X + the n-3 power of X + +X to the power of 1 + 1)
So (the nth power of x-1) is divided by (x-1) = the N-1 power of X + the N-2 power of X + the n-3 power of X + +The first power of X + 1
So LIM (x → 1) [(the nth power of x-1) divided by (x-1)]
=LIM (x → 1) (n-1 power of X + n-2 power of X + n-3 power of X + +X to the power of 1 + 1)
=n

Find the nth power of LIM = 2 + the nth power of 3, the N + 1 power of 2 + the N + 1 power of 3 = 3, and find the limit

Divide by 3 ^ n
=lim[(2/3)^n+1]/[2*(2/3)^n+3]
(2 / 3) ^ n tends to 0
So the original formula = (0 + 1) / (0 + 3) = 1 / 3

What is the limit of n's x power minus n's - x power when n approaches infinity by dividing n's x power and n's - x power?

The limit of [n ^ x-n ^ (- x)] / [n ^ x + n ^ (- x)] when n approaches infinity
The limit of formula [n ^ x-n ^ (- x)] / [n ^ x + n ^ (- x)] must be considered
When x is greater than 0
Divide the numerator and denominator of formula [n ^ x-n ^ (- x)] / [n ^ x + n ^ (- x)] by n ^ X
When n approaches infinity, its limit value is 1
When x = 0, [n ^ x-n ^ (- x)] / [n ^ x + n ^ (- x)] = 0
When x is less than 0
Divide the numerator and denominator of formula [n ^ x-n ^ (- x)] / [n ^ x + n ^ (- x)] by n ^ (- x)
[n ^ (2x) - 1] [n ^ (2x) + 1], when n approaches infinity, its limit value is - 1