Find the function limit Lim x → 0 e ^ x-x-1 / X cos x

Find the function limit Lim x → 0 e ^ x-x-1 / X cos x

LIM (e ^ x-x-1) / xcosx
x→0
=LIM (e ^ x-1) / (cosx xsinx) (it is already a fixed formula and is directly substituted into the calculation)
x→0
=(e^0-1)/(cos0-0sin0)
=(1-1)/(1-0)
=0/1
=0

Find function limit: LIM (x - > 0) (COS x) ^ (1 / x)

LIM (1-cosx under the following sign) / (1-cos following sign x) ＾ 2, X
Tends to 0 +
=LIM (1 / (1-cos follow number x)
X tends to 0 +
=+∞

Find the limit LIM (x tends to 0) (cosx) ^ (1 / x ^ 2)

x->0
lim(cosx)^(1/x^2)
=lime^(lncosx)/x^2
=e^lim(lncosx)/x^2
X - > 0 l hospital rule
lim (lncosx)/x^2
=lim-sinx/2xcosx
=lim -1/2cosx
=-1/2
So the original formula = e ^ (- 1 / 2)

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

lim(x→0)x ²/ 1-cosx
=lim(x→0)x ²/ [1-(1-2sin ² （x/2））】
=lim(x→0)x ²/ [2sin ² （x/2）】
=lim(x→0)x ²/ （x ²/ 2）
=2

Find the limit LIM (x → 0) (1-cosx) / x ^ 2 Detailed solution process

Typical infinity infinity amorphous, so first combine the pattern of 2 / x ^ 2-1 / (1-cosx) = [2-2cosx-x ^ 2] / x ^ 2 (1-cosx) with 0 / 0 up and down, and use the lower ratio to obtain the same derivative = [2sinx-2x] / [2x (1-cosx) + x ^ 2sinx] 0 / 0, Robita = [2cosx-2] / [2-2cosx + 2xsinx + x ^ 2cosx] 0 / 0

Find the limit LIM (x → 0) ((x ^ 2 + 1) ^ (1 / 2)) / (1-cosx)

When x → 0, (x ^ 2 + 1) ^ (1 / 2) → 1
1-cosx→0
So LIM (x → 0) ((x ^ 2 + 1) ^ (1 / 2)) / (1-cosx) →∞
If the question is right, it should be