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$$\lim_{x \to 0} \frac{\sin(2x) \cos(3x) \sin(4x)}{x \cos(5x) \sin(6x)}$$

There should be a way without applying L'Hopital's Rule two times.

Mike C
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1 Answers1

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Try rewriting this as

$$\frac{\sin 2x\cos 3x\sin 4x}{x\cos 5x\sin 6x} = \frac{4}{3}\cdot \frac{\sin 2x}{2x}\cdot\frac{\sin 4x}{4x}\cdot\frac{6x}{\sin 6x}\cdot\frac{\cos 3x}{\cos 5x}$$

and then taking limits.

rogerl
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