# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 22

$\dfrac{1}{2}$

#### Work Step by Step

Let $\lim\limits_{x \to \dfrac{\pi}{2}}f(x)=\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{\ln (\csc)}{(x-(\dfrac{\pi}{2}))^2}$ But $f(\dfrac{\pi}{2})=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{(-\cot x)}{(2x-\pi)}=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{\csc^2 (\dfrac{\pi}{2})}{2}=\dfrac{1}{2}$

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