# Question 2f5d1

Jan 25, 2018

${\lim}_{x \rightarrow \pi} \frac{\cos x - \sin \left(\frac{3 x}{2}\right)}{\cos x + \frac{\sin x}{2}}$

${\lim}_{x \rightarrow \pi} \frac{\cos x - \sin \left(\frac{3 x}{2}\right)}{\cos x + \frac{\sin x}{2}} = \frac{\cos \pi - \sin \left(\frac{3 \pi}{2}\right)}{\cos \pi + \frac{\sin \pi}{2}}$

$= \frac{- 1 - \sin \left(\frac{66}{14}\right)}{- 1 + 1}$

$= \frac{- 1 + 1}{- 1 + 1}$

$= \frac{0}{0}$

Apply L'Hopital's rule

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

$f \left(x\right) = \cos x - \sin \left(\frac{3 x}{2}\right)$
$g \left(x\right) = \cos x + \frac{\sin x}{2}$

Taking derivative of numerator

$\frac{d}{\mathrm{dx}} \left(\cos x - \sin \left(\frac{3 x}{2}\right)\right) = \frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) - \frac{d}{\mathrm{dx}} \left(\sin \left(\frac{3 x}{2}\right)\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$

Apply the chain rule

$\frac{d}{\mathrm{dx}} \left(\sin \left(\frac{3 x}{2}\right)\right)$

$f = \sin \left(u\right) , u = \frac{3 x}{2}$

$= \setminus \frac{d}{\mathrm{du}} \left(\setminus \sin \setminus \left(u \setminus\right) \setminus\right) \setminus \frac{\cancel{d}}{\cancel{d} \cancel{x}} \setminus \left(\setminus \frac{3 \cancel{x}}{2} \setminus\right)$

$= \cos \left(u\right) \frac{3}{2} = \left(\cos \left(\frac{3 x}{2}\right) \frac{3}{2}\right)$

And bring back first derivation

$= - \sin \left(x\right) - \left(\cos \left(\frac{3 x}{2}\right) \frac{3}{2}\right)$

Taking derivative of denominator

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right) + \sin \left(\frac{x}{2}\right)\right)$

$= \frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(\sin \left(\frac{x}{2}\right)\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$

Apply the chain rule

$f = u , u = \frac{x}{2}$

$\frac{d}{\mathrm{dx}} \left(\sin \left(\frac{x}{2}\right)\right) = \frac{d}{\mathrm{du}} \left(\sin \left(u\right)\right) \frac{\cancel{d}}{\cancel{\mathrm{dx}}} \left(\frac{\cancel{x}}{2}\right)$

$= \cos \left(u\right) \frac{1}{2}$

$= \cos \left(\frac{x}{2}\right) \frac{1}{2}$

Bring back 1st derivation

$= - \sin \left(x\right) + \cos \left(\frac{x}{2}\right) \frac{1}{2}$

Looking both the numerator and denominator i can guess its again a zero , so L'Hopital's rule again.

Numerator

$\frac{d}{\mathrm{dx}} \left(- \sin \left(x\right) - \cos \left(\frac{3 x}{2}\right)\right) = - \frac{d}{\mathrm{dx}} \sin \left(x\right) - \frac{d}{\mathrm{dx}} \cos \left(\frac{3 x}{2}\right)$

$- \frac{d}{\mathrm{dx}} \sin \left(x\right) = - \cos \left(x\right)$

d/dx (cos((3x)/2)(3)/2)) = 3/2(d)/(du)cos(u)d/dx((3x)/2) #

$= \frac{3}{2} \left(- \sin \left(u\right)\right) \frac{3}{2}$
$= \frac{3}{2} \left(- \sin \left(\frac{3 x}{2}\right)\right) \frac{3}{2}$
$= - \frac{9}{4} \sin \left(\frac{3 x}{2}\right)$

Numerator = $- \cos \left(x\right) + \frac{9}{4} \sin \left(\frac{3 x}{2}\right)$

Denominator = $- \sin \left(x\right) + \cos \left(\frac{x}{2}\right) \frac{1}{2}$
$\frac{d}{\mathrm{dx}} \left(- \sin \left(x\right) + \cos \left(\frac{x}{2}\right) \frac{1}{2}\right) = \frac{d}{\mathrm{dx}} \left(- \sin \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(\cos \left(\frac{x}{2}\right) \frac{1}{2}\right)$

$= - \cos \left(x\right) + \frac{1}{2} \frac{d}{\mathrm{du}} \cos \left(u\right) \frac{d}{\mathrm{dx}} \frac{x}{2}$

$= - \cos \left(x\right) + \frac{1}{2} \left(- \sin \left(\frac{x}{2}\right)\right) \frac{1}{2}$

$\sin \left(x\right) = \frac{1}{\csc \left(x\right)}$

$= - \cos \left(x\right) + \frac{1}{\csc \left(\frac{x}{2}\right) \cdot 2 \cdot 2}$

$= - \cos \left(x\right) + \frac{1}{4 \csc \left(\frac{x}{2}\right)}$

$= - \cos \left(x\right) + \frac{- \sin \left(\frac{x}{2}\right)}{4}$

$= - \cos \left(x\right) - \frac{\sin \left(\frac{x}{2}\right)}{4}$

Now plug in $\pi$ instead of $x$

Numerator
$- \cos \left(x\right) + \frac{9}{4} \sin \left(\frac{3 x}{2}\right) = - \cos \left(\pi\right) + \frac{9}{4} \sin \left(\frac{3 \pi}{2}\right)$

$\sin \left(x\right) = \cos \left(\frac{\pi}{2} - 3 \frac{\pi}{2}\right)$

$= - \cos \left(\pi\right) + \frac{9}{4} \cos \left(\frac{\pi}{2} - \frac{3 \pi}{2}\right)$

$= - \cos \left(\pi\right) + \frac{9}{4} \cos \left(\frac{{\cancel{- 2}}^{- 1} \pi}{\cancel{2}}\right)$

$- \cos \left(\pi\right) + \frac{9}{4} \cos \left(- \pi\right)$

Remember cos(-x) = cos(x)

$= - \cos \left(\pi\right) + \frac{9}{4} \cos \left(\pi\right)$

$= \cos \left(\pi\right) \left(- 1 + \frac{9}{4}\right)$

$= \cos \left(\pi\right) \left(\frac{- 4 + 9}{4}\right)$

$= - 1 \cdot \frac{5}{4}$

$= - \frac{5}{4}$

Denominator

$- \cos \left(x\right) - \frac{\sin \left(\frac{x}{2}\right)}{4}$

$= - \cos \left(\pi\right) - \frac{\sin \left(\frac{\pi}{2}\right)}{4}$

$- 1 \left(- 1\right) - \frac{1}{4}$

$\frac{1 \cdot 4 - 1}{4}$

$= \frac{3}{4}$

So ${\lim}_{x \rightarrow \pi} \frac{\cos x - \sin \left(\frac{3 x}{2}\right)}{\cos x + \frac{\sin x}{2}} = - \frac{5}{4} \cdot \frac{4}{3}$

$= - \frac{5}{3}$