#38 Measure Your Calibre

Set $\displaystyle x\to x+\pi$ ,

$\displaystyle \begin{aligned} L &=\lim_{x\to 0} \dfrac{\tan(\pi \cos^2 x)}{x^2} \\ &=-\lim_{x\to 0} \dfrac{\tan(\pi-\pi\cos^2 x)}{x^2} \\ &=-\lim_{x\to 0} \dfrac{\tan(\pi\sin^2 x)}{\sin^2 x}\times \dfrac{\sin^2 x}{x^2} \\ &=-\pi\times\left(\lim_{x\to 0}\dfrac{\tan(\pi\sin^2 x)}{\pi\sin^2 x}\right)\times\left(\lim_{x\to 0}\dfrac{\sin x}{x}\right)^2 \\& =-\pi \end{aligned}$

which makes $\dfrac{L}{\pi}=-1$

Plug $x = \pi +h$ $\begin{aligned}L & = \lim_{x\to \pi} \dfrac{\tan \,(\pi \cos^2 x)}{\,(\pi-x)^2} \\& = \lim_{h\to 0} \dfrac{\tan\,(\pi \cos ^2 h)}{h^2} \phantom{cccccc}{\color{#3D99F6} \dfrac{0}{0} \text{ form, As }x\to \pi , h \to 0} \\& = \lim_{h\to 0}\dfrac{\sec ^2\,(\pi \cos^2 h)\,(-\pi \sin 2h)}{2h} = -\pi \end{aligned}$ Therefore, $\dfrac{L}{\pi} = \boxed{-1}$ .