Solving LIM (x → + ∞) e ^ X / x ^ n (n is a natural number) with lobida

Solving LIM (x → + ∞) e ^ X / x ^ n (n is a natural number) with lobida

The final result is e ^ X / N! (x - + ∞), so the result is + ∞

lim(cosx+cos2x+ cos3x+… +Cos nx-n) / (cosx-1) (x tends to 0)
2, 3, N and so on are powers, not 2 times X

consider
1+2+3+..+n = n(n+1)/2
n^2 = n(n+1)-n
= (1/3) [n(n+1)(n+2)-(n-1)n(n+1)] -n
1^2+2^2+3^2+..+n^2
=(1/3)n(n+1)(n+2) -n(n+1)/2
=(1/6)n(n+1)(2n+1)
lim(x->0) (cosx+cos2x+ cos3x+… + cosnx-n)/(cosx-1) (0/0)
= lim(x->0) (sinx+2sin2x+ 3sin3x+… + nsinnx)/(sinx) (0/0)
=lim(x->0) (cosx+2^2cos2x+ 3^3cos3x+… + n^2cosnx)/(cosx)
=1^2+2^2+..+n^2
=(1/6)n(n+1)(2n+1)

Find LIM (X & # 178; + X √ (X & # 178; + 2) x → - ∞

-1. See figure

lim(n-∞)﹙x²／x²-1﹚
lim(n-∞)[x²／x²-1﹚]^x

lim[x²／x²-1﹚]^x
=lim[1+1/(x^2-1)]^x
=lim[1+1/(x^2-1)]^{(x^2-1)]*[x^2/(x^2-1)]}
=lim[1+1/(x^2-1)]^(x^2-1)]
=E (1)
∴A1B1=(-1,1,0) A1A=(0,0,√2) C1D=(1/2,1/2,0)
∴A1B1·C1D=0 A1A·C1D=0
∴A1B1⊥C1D A1A⊥C1D
Moreover, A1B1 ∈ plane aa1b1b and A1A ∈ plane aa1b1b and A1A and A1B1 are not parallel
〈 C1d ⊥ plane aa1b1b
(2)、∵AB1=(-1,1,-√2)
∴AB1·C1D=0
∴AB1⊥C1D
As long as Ab1 ⊥ DF, there will be Ab1 ⊥ plane c1df
And ∵ DF = (- 1 / 2,1 / 2, z)
∴AB1·DF=1-z√2
When z = 1 / √ 2, Ab1 ⊥ DF
That is: when the coordinate of F point is (0,1,1 / √ 2), that is, the midpoint of BB1, it will make Ab1 ⊥ plane c1df

2 / x times of LIM x → 0 (x + 3x) = 6 times of a.1 B.E. & # 178; C.E. & # 179; D.E

There is a problem with the title and the result is 0

Find LIM (1-cosx) / x ^ 2
Find LIM (1-cosx) / x ^ 2
X→0

A:
lim(x→0) (1-cosx)/x²
=lim(x→0) 2sin²(x/2)/[4*(x/2)²]
=lim(t→0) (1/2）(sint/t)²
=1/2

lim(x->∞)[ xsin(2/x)+(cosx^2)/(x^2)]

Equivalent substitution sin1 / x ~ 1 / x, cosx ~ 1-x ^ 2

x→∞lim(x-cosx)/x=?

∵lim(x→∞)［（x－cosx）/x］＝lim(x→∞)（1－cosx/x）＝1－lim(x→∞)（cosx/x）.
It is obvious that cosx is bounded, and ﹥ LIM (x →∞) (cosx / x) = 0, ﹥ LIM (x →∞) [(x-cosx) / x] = 1

lim（e＾x）cosx= x→+∞

x→+∞
Then e ^ x → + ∞
And cosx oscillates at [- 1,1], that is bounded
So e ^ x * cosx →∞
The limit does not exist

LIM (x - > x0) f (x) / X limit exists and f (x) is continuous at x0
Is f (x) derivable at x0?

Whether f (x) is differentiable at x0, that is, whether LIM (x - > x0) (f (x) - f (x0)) / (x-x0) exists,
Since f (x) is continuous at x0, that is, when X - > x0, f (x) - > F (x0), the law of lobita is used,
lim(x->x0)(f(x)-f(x0))/(x-x0)=lim(x->x0)f'(x)=lim(x->x0)(f(x)-f(0))/(x-0)=lim(x->x0)f(x)/x.
So it can be derived. I wonder if there is a condition that f (0) = 0 in the title?