#38 Measure Your Calibre

Calculus Level 4

L = lim x π tan ( π cos 2 x ) ( x π ) 2 \large L = \lim_{x \to \pi} \frac{\tan(\pi \cos^2x)}{(x-\pi)^2} .

Find L π \dfrac{L}{\pi}


The answer is -1.

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2 solutions

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

L = lim x 0 tan ( π cos 2 x ) x 2 = lim x 0 tan ( π π cos 2 x ) x 2 = lim x 0 tan ( π sin 2 x ) sin 2 x × sin 2 x x 2 = π × ( lim x 0 tan ( π sin 2 x ) π sin 2 x ) × ( lim x 0 sin x x ) 2 = π \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 L π = 1 \dfrac{L}{\pi}=-1

Naren Bhandari
Aug 10, 2018

Plug x = π + h x = \pi +h L = lim x π tan ( π cos 2 x ) ( π x ) 2 = lim h 0 tan ( π cos 2 h ) h 2 c c c c c c 0 0 form, As x π , h 0 = lim h 0 sec 2 ( π cos 2 h ) ( π sin 2 h ) 2 h = π \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, L π = 1 \dfrac{L}{\pi} = \boxed{-1} .

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