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Question Number 148992 by mathlove last updated on 02/Aug/21

Answered by gsk2684 last updated on 02/Aug/21

$$\approx \underset{x\to 0}{\lim }\frac{(\sin x+\frac{\sin {}^{2}x}{2!}+..)-(\sin 3x+\frac{\sin {}^{2}3x}{2!}+..)}{2x}$$$$=\frac{1}{2}\{(1+0+...)-(3+0+..)\}$$$$=-1$$

Answered by iloveisrael last updated on 02/Aug/21

$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\sin \mathrm{x}}-{\mathrm{e}}^{\sin 3\mathrm{x}}}{\sin 2\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\cos \mathrm{x}{\mathrm{e}}^{\sin \mathrm{x}}-3\cos 3\mathrm{x}{\mathrm{e}}^{\sin \mathrm{x}}}{2\cos 2\mathrm{x}}$$$$=\frac{1-3}{2}=-1$$

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