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\[\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin 2x}} - {e^{\sin x}}}}{{\sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{\sin 2x}} - 1}}{{\sin x}} - \frac{{{e^{\sin x}} - 1}}{{\sin x}}} \right)\]\[ = \mathop {\lim }\limits_{x \to 0} \left( {2\cos x.\frac{{{e^{\sin 2x}} - 1}}{{\sin 2x}} - \frac{{{e^{\sin x}} - 1}}{{\sin x}}} \right) = 1\]\[\left( {do{\rm{ }} \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1} \right)\]
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