Limx → 0 ∫ cosx is the limit of negative T square / x square of the lower limit and 1 is the upper limit E. (detailed process.) thank you

Limx → 0 ∫ cosx is the limit of negative T square / x square of the lower limit and 1 is the upper limit E. (detailed process.) thank you

limx→0∫（cosx,1)e^(-t^2)dt／x^2
=limx→0e^(-(cosx)^2)sinx／2x
=limx→0e^(-(cosx)^2)／2
=1/(2e)

How to find the limit of limx →∞ (x + cosx + 1 / x + SiNx + 2)?

Cosx and SiNx are bounded functions
So when x →∞
They can be omitted, so the limit is 1

Find the limit value of limx approaching 0 1-radical cosx / x ^ 2

LIM (1-root cosx / x ^ 2) = LIM ((x ^ 2-root cosx) / x ^ 2)
Derivation of numerator denominator by Robita's law = LIM ((2x + 1 / 2sinx / radical cosx) / 2x)
=LIM (2 + 1 / 2 (cosx root cosx SiNx / root cosx) / cosx) / 2)
=5/4

Find the limit limx → infinity (SiNx ^ 2-x) / [(cosx) ^ 2-x] My head is big. The answer in the book seems to be - 1.

When x tends to infinity. Due to - 1=

Limx tends to 0 (SiNx / x) ^ 1 / (1-cosx) lobida's law

Robita's law formula Limu ^ v = e ^ (limvlnu) [applicable to finding 1 ^ infinite, infinite ^ 0,0 ^ 0 type limit] where u = SiNx / x, v = 1 / (1-cosx) limvlnu = Lim [ln (SiNx / x)] / (1-cosx) molecular derivation of Robita's law [ln (SiNx / x)] '= (x / SiNx) [(xcosx SiNx) / X ²]...

Find the limit limx → o (cosinx cosx) / (x2 (1-cosx)) =?

When the common equivalent infinitesimal x → 0, 1-cosx ~ x ²/ 2. When SiNx → 0, cosinx ~ 1-sin ² X / 2, then limx → o (cosinx cosx) / (x) ² (1-cosx))=limx→o(1-sin ² x/2-cosx)/(x ² (1-cosx))=limx→o(1-cosx) ²/ (2x ² (1-cosx)...