lim (1/(2x) -1/tan(2x))

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac 1{2x} - \frac 1{\tan 2x} \right) & \small \blue{\text{Let }u = 2x} \\ & = \lim_{u \to 0} \left(\frac 1u - \frac 1{\tan u} \right) \\ & = \lim_{u \to 0} \left(\frac 1u - \frac {\cos u}{\sin u} \right) \\ & = \lim_{u \to 0} \frac {\sin u - u \cos u}{u \sin u} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{u \to 0} \frac {\cos u - \cos u + u \sin u}{\sin u + u \cos u} & \small \blue{\text{Differentiate up and down w.r.t. }u} \\ & = \lim_{u \to 0} \frac {u \sin u}{\sin u + u \cos u} & \small \blue{\text{A 0/0 case again}} \\ & = \lim_{u \to 0} \frac {\sin u + u \cos u}{\cos u + \cos u - u \sin u} & \small \blue{\text{Differentiate up and down w.r.t. }u} \\ & = \frac 02 = \boxed 0 \end{aligned}$

Reference: L'Hôpital's rule

The given expression can be rewritten as

$\displaystyle \lim_{x\to 0} \dfrac {1-\frac{2x}{\tan 2x}}{2x}$

Under the limit, this assumes the $\dfrac 00$ form, and hence applying L'Hospital's rule repeatedly we get the result $\boxed 0$ .