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

Please write the steps out. The answer should be -√3 right?

Put z = θ–π/3 and let L = lim [θ→π/3] {2cos(θ)−1}/(θ–π/3)

2cos(θ) – 1 = 2cos(z)cos(π/3) – 2sin(z)sin(π/3) − 1

= cos(z) − 1 − √3.sin(z) = −2sin²(z/2) − √3.sin(z)

Hence L = lim [z→ 0] −2sin²(z/2)/z − √3.sin(z)/z

= lim [z→0] −{sin(z/2)/(z/2)}sin(z/2) − √3{sin(z)/z}

As x→0 sin(x)/x → 1 (this is from the basic geometry of a circle)

Hence L = −1*0 − √3*1 = −√3

I am assuming that you are required to not use the Hospital :)
#1 With trig functions you want the limit to go to 0 since most likely you will be looking to use lim α➝0 (sinα/α)=1 . So make a change of variable: α=θ-π/3
#2 Use formula cos(x+y)=cos(x)cos(y)-sin(x)sin(y) and substitute values for cos(π/3)=1/2 and sin(π/3)=√3/2.

2cosθ -1=2cos(α+π/3)-1 =2(cos(α)cos(π/3)-sin(α)sin(π/3))-1=
(cont.) = cos(α)-√3sin(α)-1.
(2cosθ -1)/(θ - π⁄₃)=(cos(α)-√3sin(α)-1)/α .
Take the limit
lim α➝0 (cos(α)-√3sin(α)-1)/α =lim α➝0 (1-√3sin(α)-1)/α =-√3 lim α➝0 (sinα/α)= -√3.

Let f(x) = 2 cos(t)
f(pi/3) = 2 cos(pi/3) = 2 sin(pi/6) = 1

The limit is lim_(t --> pi/3) (f(t) - f(pi/3)/(t - pi/3) = f'(pi/3)

This is just the definition of the derivative!

In this case f'(t) = - 2 sin(t); f'(pi/3) = - 2 sin(pi/3) = - 2 sqrt(3)/2 = - sqrt(3)


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Note: The definition of the derivative is the basis of L'Hopital's rule.
In fact it is the only case known to him. Moreover he did not invent it, Bernoulli did.
L'Hopital credits Bernoulli. Here''s what is in L'H 's book:

(f(x) - f(a))/(g(x) - g(a)) =

= [(f(x) - f(a))/(x-a)] [(x-a)/(g(x) - g(a)) ] --> f'(a)/g'(a) as x --> a

The assumption is that g'(a) is not 0.

Stop trying to make life complicated, L'Hôpital's rule works here so you should use it.