Limit Exercise for Trigonometric Functions

Compute, $\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin 2x}{\sin 3x}$

Solution:
we have $\frac{\sin 2x}{\sin 3x}=\frac{2\sin x\cos x}{\sin \left( x+2x \right)}$But $\sin \left( x+2x \right)=\cos \left( 2x \right)\sin x+\cos \left( x \right)\sin \left( 2x \right)$

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

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

Thus $\frac{\sin 2x}{\sin 3x}=\frac{2\sin x\cos x}{\sin x\left( 4{{\cos }^{2}}x-1 \right)}=\frac{2\cos x}{4{{\cos }^{2}}x-1}=\frac{2\cos x}{2\left( 1+\cos 2x \right)-1}=\frac{2\cos x}{2\cos 2x+1}$

So $\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin 2x}{\sin 3x}=\underset{x\to \pi }{\mathop{\lim }}\,\frac{2\cos x}{2\cos 2x+1}=-\frac{2}{3}$