# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 18

$3$

#### Work Step by Step

Consider: $\lim\limits_{\theta \to \dfrac{-\pi}{3}}f(\theta)=\lim\limits_{ \theta \to \dfrac{-\pi}{3}}\dfrac{3 \theta +\pi}{\sin (\theta+\dfrac{\pi}{3})}$ Need to check that the limit has an indeterminate form. Thus, $f(\dfrac{-\pi}{3})=\dfrac{3 (\dfrac{-\pi}{3}) +\pi}{\sin ((\dfrac{-\pi}{3})+\dfrac{\pi}{3})}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{\theta \to \dfrac{-\pi}{3}}\dfrac{3}{\cos (\theta+\dfrac{\pi}{3})}=\dfrac{3}{\cos (\dfrac{-\pi}{3}+\dfrac{\pi}{3})}=\dfrac{3}{\cos 0}=3$

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