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Question

Evaluate the following limits. Use I Hópital's Rule when it is convenient and applicable. limxπ/2tanx3/(2xπ)\lim _{x \rightarrow \pi / 2^{-}} \frac{\tan x}{3 /(2 x-\pi)}

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Here, we are dealing with the intermediate form \dfrac{\infty}{\infty} and in order to calculate this limit we need to apply L'Hopital Rule.

limx(π2)tan(x)32xπ=13limx(π2)sec2(x)2(2xπ)2=13limx(π2)(sec2(x)(2xπ)22)=16limx(π2)(sec2(x)(2xπ)2)=16limx(π2)((π+2x)21sec2(x))=16limx(π2)(4(π+2x)sin(2x))=16limx(π2)(82cos(2x))=16limx(π2)(4cos(2x))=16(4cos(2π2))=23\begin{align*} \lim\limits_{x \to (\frac{\pi}{2})^-} \dfrac{\tan(x)}{\dfrac{3}{2x-\pi}}&= \dfrac{1}{3}\lim\limits_{x \to (\frac{\pi}{2})^-} \dfrac{\sec^2(x)}{-\dfrac{2}{(2x-\pi)^2}}\\[7pt] &=\dfrac{1}{3}\lim\limits_{x \to (\frac{\pi}{2})^-} \left(-\dfrac{\sec^2(x)(2x-\pi)^2}{2} \right)\\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \sec^2(x)(2x-\pi)^2 \right)\\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{(-\pi+2x)^2}{\dfrac{1}{\sec^2(x)}} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{4(-\pi +2x)}{-\sin(2x)} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{-8}{2\cos(2x)} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{-4}{\cos(2x)} \right) \\[7pt] &=-\dfrac{1}{6} \left( \dfrac{-4}{\cos(2\cdot \dfrac{\pi}{2})} \right) \\[7pt] &=-\dfrac{2}{3} \end{align*}

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