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Question Number 215343 by RoseAli last updated on 03/Jan/25

$$\underset{x\to 0}{\lim }\frac{\sin 3x\cos 5x}{x^2}$$

Commented by Frix last updated on 04/Jan/25

$$\mathrm{Does}\mathrm{not}\mathrm{exist}.$$

Answered by MrGaster last updated on 04/Jan/25

$$\underset{x\to 0}{\lim }\frac{\sin 3x\cos 5x}{x^2}$$$$\underset{x\to 0}{\lim }(\frac{\sin 3x}{3x}\cdot \frac{3\cos 5x}{2x})$$$$=\underset{x\to 0}{\lim }\left(\frac{\sin 3x}{3x}\right)\cdot \underset{x\to 0}{\lim }\left(\frac{3\cos 5x}{2x}\right)$$$$=1\cdot \underset{x\to 0}{\lim }\left(\frac{3\cos 5x}{2x}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{3\cos 5x}{2x}\right)$$$$=\underset{x\to 0}{\lim }(\frac{3\cos 5x}{2}\cdot \frac{1}{x})$$$$=\frac{3}{2}\underset{x\to 0}{\lim }\left(\frac{\cos 5x}{x}\right)$$$$=\frac{3}{2}\cdot \underset{x\to 0}{\lim }\left(\frac{\cos 5x-\cos 5\cdot 0}{x-0}\right)$$$$=\frac{3}{2}\cdot (-5\sin 5\cdot 0)$$$$=\frac{3}{2}\cdot (-5\cdot 0)$$$$=0\mathrm{or}-15$$

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