# What is the correct option of the following question?

Feb 9, 2018

$\setminus$

$\text{Answer is:" \qquad \quad "Choice C.}$

#### Explanation:

$\setminus$

$\text{We compute, and use L'Hosptal's Rule:}$

${\lim}_{x \rightarrow 0} \frac{1 - \cos \left(x\right) \cos \left(2 x\right) \cos \left(3 x\right)}{S {\in}^{2} \left(2 x\right)} \setminus q \quad \setminus q \quad \frac{0}{0} , \text{use L'Hospital's Rule}$

$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} \frac{\left[1 - \cos \left(x\right) \cos \left(2 x\right) \cos \left(3 x\right)\right] '}{\left[S {\in}^{2} \left(2 x\right)\right] '}$

$\setminus$
$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} - \frac{\left[\cos \left(x\right)\right] ' \cos \left(2 x\right) \cos \left(3 x\right) + \cos \left(x\right) \left[\cos \left(2 x\right)\right] ' \cos \left(3 x\right) + \cos \left(x\right) \cos \left(2 x\right) \left[\cos \left(3 x\right)\right] '}{2 \setminus \cdot \sin \left(2 x\right) \setminus \cdot 2}$

$\setminus$
$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} - \frac{\left[- \sin \left(x\right)\right] \cos \left(2 x\right) \cos \left(3 x\right) + \cos \left(x\right) \left[- 2 \sin \left(2 x\right)\right] \cos \left(3 x\right) + \cos \left(x\right) \cos \left(2 x\right) \left[- 3 \sin \left(3 x\right)\right]}{2 \setminus \cdot \sin \left(2 x\right) \setminus \cdot 2}$

$\setminus$
$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} \frac{\sin \left(x\right) \cos \left(2 x\right) \cos \left(3 x\right) + 2 \cos \left(x\right) \sin \left(2 x\right) \cos \left(3 x\right) + 3 \cos \left(x\right) \cos \left(2 x\right) \sin \left(3 x\right)}{2 \setminus \cdot \sin \left(2 x\right) \setminus \cdot 2}$

$\setminus$
$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} \frac{\sin \left(x\right) \cos \left(2 x\right) \cos \left(3 x\right) + 2 \cos \left(x\right) \left(2 \sin \left(x\right) \cos \left(x\right)\right) \cos \left(3 x\right) + 3 \cos \left(x\right) \cos \left(2 x\right) \left(3 {\cos}^{2} \left(x\right) \sin \left(x\right) - {\sin}^{3} \left(x\right)\right)}{4 \setminus \cdot \left(2 \sin \left(x\right) \cos \left(x\right)\right)}$

$\setminus$
$\text{[Remove factor of" \ sin(x) \ "from top and bottom:]}$

$\setminus q \quad \setminus q \quad = \setminus {\lim}_{x \rightarrow 0} \frac{\cos \left(2 x\right) \cos \left(3 x\right) + 2 \cos \left(x\right) \left(2 \cos \left(x\right)\right) \cos \left(3 x\right) + 3 \cos \left(x\right) \cos \left(2 x\right) \left(3 {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right)}{4 \setminus \cdot \left(2 \cos \left(x\right)\right)}$

$\setminus$
$\text{[Substitute" \ x=0 ":]}$

$\setminus q \quad \setminus q \quad = \setminus \frac{1 \setminus \cdot 1 + 2 \setminus \cdot 1 \left(2 \setminus \cdot 1\right) \setminus \cdot 1 + 3 \setminus \cdot 1 \setminus \cdot 1 \left(3 \setminus \cdot {1}^{2} - {0}^{2}\right)}{4 \setminus \cdot \left(2 \setminus \cdot 1\right)}$

$\setminus q \quad \setminus q \quad = \setminus \frac{1 + 4 + 9}{8} \setminus = \setminus \frac{14}{8}$

$\setminus q \quad \setminus q \quad = \frac{7}{4.}$

$\setminus$

$\text{So, following the statement of the problem, the answer is:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{Choice C.} \setminus \quad \setminus \quad \setminus \quad \setminus \square$