## Related questions with answers

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

Solution

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

$\begin{align*} \lim\limits_{x \to (\frac{\pi}{2})^-} \dfrac{\tan(x)}{\dfrac{3}{2x-\pi}}&= \dfrac{1}{3}\lim\limits_{x \to (\frac{\pi}{2})^-} \dfrac{\sec^2(x)}{-\dfrac{2}{(2x-\pi)^2}}\\[7pt] &=\dfrac{1}{3}\lim\limits_{x \to (\frac{\pi}{2})^-} \left(-\dfrac{\sec^2(x)(2x-\pi)^2}{2} \right)\\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \sec^2(x)(2x-\pi)^2 \right)\\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{(-\pi+2x)^2}{\dfrac{1}{\sec^2(x)}} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{4(-\pi +2x)}{-\sin(2x)} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{-8}{2\cos(2x)} \right) \\[7pt] &=-\dfrac{1}{6}\lim\limits_{x \to (\frac{\pi}{2})^-} \left( \dfrac{-4}{\cos(2x)} \right) \\[7pt] &=-\dfrac{1}{6} \left( \dfrac{-4}{\cos(2\cdot \dfrac{\pi}{2})} \right) \\[7pt] &=-\dfrac{2}{3} \end{align*}$

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