How am i to solve limθ➝π⁄₃ (2cosθ -1)/(θ - π⁄₃) without using the L'Hôpital's Rule

Sub t=θ-π⁄₃, (2cosθ -1)/(θ - π⁄₃)=(2cos(t+π⁄₃)-1)/t
=[2costcosπ⁄₃-2sintsinπ⁄₃]/t=[cost-√3s…
Using series expansions of cost and sint,
...=[1-t^2/2!+..-√3(t-t^3/3!+..)-1]/t
=[-√3t -t^2/2+...]/t
=-√3 -t/2+...
And as t->0 this has limit -√3.
So right!.
Of course it is easier using the H R.

put x-pi/3=h, x=>pi/3 , h=>0,

h=>0 [2cos(pi/3+h)-1] [2(cospi/3cosh-sinpi/3sinh)-1] [2(1/2cosh-root3/2sinh)-1]
----------------------- = h=>0------------------------------------… =h=>0 ---------------------------------
h h h
[cosh-root3sinh-1] [-root3sinh-(1-cosh)] -root3sinh 2sin^2h/2
h=>0--------------------------- =h=>0------------------------------=h=>0 -------------- - -------------- =--root3 -0= --root3
h h h h