# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 50

$\dfrac{1}{2}$

#### Work Step by Step

Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0} \dfrac{3 \cos 3x-3+2x}{(\cos 2x)\sin 2x+(2\cos 2x) \sin x}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0} \dfrac{2-9 \sin 3x}{-5\sin 2x\sin x+4 \cos 2x \cos x}=\dfrac{2-9 \sin 0}{-5\sin 0\sin 0+4 \cos 0 \cos 0}$ or, $\dfrac{2}{0+4}=\dfrac{1}{2}$

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