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Find the limit of the given function

by RoRi

Find the value of the following two limits

    \[ \lim_{x \to 0} \frac{(\sin (4x))(\sin (3x))}{x \sin (2x)}, \quad \text{and} \quad \lim_{x \to \frac{\pi}{2}} \frac{(\sin (4x))(\sin (3x))}{x \sin (2x)}. \]

For the limit as x \to 0 we the identity \sin (4x) = \sin (2(2x)) = 2 \sin (2x) \cos (2x) and the limit \lim_{x \to 0} \frac{\sin x}{x} = 1,

    \begin{align*} \lim_{x \to 0} \frac{(\sin (4x))(\sin (3x))}{x \sin (2x)} &= \lim_{x \to 0} \left( \frac{2 \sin (2x) \cos (2x)}{\sin (2x)}\cdot \frac{\sin (3x)}{x} \right) \\[9pt] &= \lim_{x \to 0}\left( 2 \cos (2x) \cdot 3 \cdot \frac{\sin (3x)}{3x} \right) \\[9pt] &= 2 \cdot 3 \cdot 1 = 6. \end{align*}

For the limit as x \to \frac{\pi}{2} we again use the identity \sin (4x) = 2 \sin (2x) \cos (2x),

    \begin{align*} \lim_{x \to \frac{\pi}{2}} \frac{(\sin (4x))(\sin (3x))}{x \sin (2x)} &= \lim_{x \to \frac{\pi}{2}} \left( 2 \cos (2x) \cdot \frac{\sin (3x)}{x} \right) \\[9pt] &= (-2) \cdot \left( \frac{-2}{\pi}\right) \\[9pt] &= \frac{4}{\pi}. \end{align*}

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