Is there a LIMIT to the amount of limits problems?

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

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Reference: L'Hôpital's rule

With out L 'Hôpital's rule

$Y= x \tan (x) - (π/2) \tan (x)$

$=(x - π/2) \tan (x)$

$=(x - π /2) \cot (π /2 - x)$

$=[(x - π /2) × \cos (π /2) ]/(\sin (π /2 - x))$

As $x$ reaches to $(π /2)$
$\sin(π /2-x)=(π /2 - x)$

So the Y becomes $(x-π /2)/(π /2-x) ×\cos (π /2 - x)$

$=-1×\cos (z)$ (where $z$ reaches to zero)

= $-1×1 =-1$