Lim x tends to 0 and the limit of (1 + x) to the power of 1 / X - E / X

Lim x tends to 0 and the limit of (1 + x) to the power of 1 / X - E / X

lim(x->0) [(1+x)^(1/x)-e]/x=lim(x->0) {e^[(1/x)ln(1+x)]-e}/x=lim(x->0) (1+x)^(1/x)*[1/x(1+x)-ln(1+x)/x^2]=lim(x->0) (1+x)^(1/x)* lim(x->0) [x-(1+x)ln(1+x)]/(1+x)x^2=e*lim(x->0) [1-ln(1+x)-1]/[x^2+2x(1...

Find the limit LIM (SiNx / x) ^ 1 / (1-cosx) [x → 0] sinx 1 ----The limit of - th power in [x → 0] x 1-cosx

The answer should be 1,
Because I don't quite understand what you wrote. I understand it like this: find the limit of SiNx / X to the power of 1 / (1-cosx) when x → 0

To find the limit Lim x tends to π / 2, (SiNx) of (1 / cosx) ^ 2 Find the limit Lim x tends to π / 2, (SiNx) to the power of (1 / cosx ^ 2)

Let f (x) = (SiNx) ^ (1 / cos? X), LN f (x) = (1 / cos? X) ln (SiNx) LIM (x - > π / 2) ln f (x) = LIM (x - > π / 2) ln (SiNx) / cos? X = LIM (T - > 0) ln (cost) / sin? T such that t = π / 2 - x, T - > 0 = LIM (T - > 0) ln (1 +...)

Find the limit of LIM (SiNx + cosx) ^ x when x tends to 0

When x → 0
sinx→0
cosx→1
x→0
Then the original formula = 1 ^ 0 = 1
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LIM (1-cosx + SiNx) limit (x tends to 0) It's the equivalent of infinitesimal X

lim(1-cosx+sinx)
=lim(2sin^x/2+sinx)
=lim[2(x/2)^2+x]
=0

Find the limit LIM (x → Ω / 2) = [SiNx] ^ 1 / (cosx) ^ 2

When x tends to π / 2, Lim SiNx = 1, LIM (cosx) ^ 2 = 0 ((cosx) ^ 2 > 0)
So Lim [SiNx / (cosx) ^ 2] = + ∞

Limit: LIM (x-0) (x * cosx SiNx) / (x ^ 3) =? Why can't we reduce it to (xcosx-x) / (x ^ 3), and then reduce x to get (cosx-1) / (x ^ 2), and then use lobida's law to get (- SiNx) / (2x) = - 1 / 2

Because the equivalent infinitesimal can not be used in addition and subtraction, only in multiplication and division
lim(x-0)（x*cosx-sinx）/(x^3)=lim(x-0)（cosx-x*sinx-cosx）/(3x^2)=-1/3

Find the limit LIM (x tends to 0) SiNx / √ (1-cosx)

lim(x->0) sinx/√（1-cosx）
=lim(x->0) sinx/ sin(x/2) (0/0)
=lim(x->0) cosx/[ (1/2)cos(x/2)]
=2

Limit LIM (x →∞) {x + SiNx / x + cosx}

The answer is 1. The goal of this question is not to allow the use of the law of rabida, because it does not conform to the conditions of the law, but you can understand that x is super large, so the latter two sine and cosine almost have no effect, and the effect of the limit weakens it to 0. The result can be imagined, equivalent to X / x = 1

Urgent! Find LIM (1-cosx) ^ (2secx), X → π / 2 limit

The original formula = LIM (x - > π / 2) {[1 + (- cosx)] ^ [(1 / (- cosx)) (- 2)]}
=【lim(x->π/2){[1+(-cosx)]^[1/(-cosx)]}】^(-2)
=E ^ (- 2) (application important limit LIM (Z - > 0) [(1 + Z) ^ (1 / z)] = e)
=1/e².