# What is lim_(x to pi) (sin3x)/(sin2x) ?

Jan 21, 2018

$- \frac{3}{2}$

#### Explanation:

${\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} \to \frac{0}{0}$

So we can use L'Hopital's rule:

$\frac{d}{\mathrm{dx}} \sin 3 x = 3 \cos 3 x$

$\frac{d}{\mathrm{dx}} \sin 2 x = 2 \cos 2 x$

$\therefore {\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} = {\lim}_{x \to \pi} \frac{3 \cos 3 x}{2 \cos 2 x}$

$= \frac{3 \cos \left(3 \pi\right)}{2 \cos \left(2 \pi\right)} = \frac{3 \cdot \left(- 1\right)}{2 \cdot 1} = - \frac{3}{2}$

This is in agreement with the graph of the function:

graph{(sin(3x))/(sin(2x)) [-7.31, 12.69, -5.16, 4.84]}

Clicking on the graph at the right location, it can be seen that at $x = \pi$, $y = - \frac{3}{2} = - 1.5$

Jan 21, 2018

${\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} = - \frac{3}{2}$

#### Explanation:

${\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} {=}_{D . L . H}^{\left(\frac{0}{0}\right)}$

${\lim}_{x \to \pi} \frac{\left(3 x\right) ' \cos 3 x}{\left(2 x\right) ' \cos 2 x}$ $=$

$\frac{3}{2} {\lim}_{x \to \pi} \frac{\cos 3 x}{\cos 2 x}$ $=$

$\frac{3}{2} \cdot \frac{\cos 3 \pi}{\cos 2 \pi}$ $=$

$- \frac{3}{2}$

because
$\cos 3 \pi = - 1$
$\cos 2 \pi = 1$

Jan 21, 2018

${\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} = - \frac{3}{2}$

#### Explanation:

Here's a method without using l'Hopital's rule. I personally feel this method to be more challenging and fun.

$\sin 3 x = \sin \left(2 x + x\right) = \sin 2 x \cos x + \cos 2 x \sin x =$

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

$2 \sin x {\cos}^{2} x + {\cos}^{2} x \sin x - {\sin}^{3} x =$

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

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

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

$\sin x \left(2 \cos 2 x + 1\right)$

$\sin 2 x = 2 \sin x \cos x$

So $\frac{\sin 3 x}{\sin 2 x} = \frac{\sin x \left(2 \cos 2 x + 1\right)}{2 \sin x \cos x} = \frac{2 \cos 2 x + 1}{2 \cos x}$

And ${\lim}_{x \to \pi} \frac{\sin 3 x}{\sin 2 x} = {\lim}_{x \to \pi} \frac{2 \cos 2 x + 1}{2 \cos x} = \frac{2 \cos 2 \pi + 1}{2 \cos \pi} = - \frac{3}{2}$