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Question Number 184378 by Ml last updated on 05/Jan/23

$$\underset{x\to \mathrm{\infty }}{\lim }{(-\ln \frac{21}{10})}^{2\mathrm{x}}=?$$

Answered by SEKRET last updated on 05/Jan/23

$$0$$$$$$

Commented by Ml last updated on 05/Jan/23

$$\mathrm{solution}$$

Answered by JDamian last updated on 05/Jan/23

$$\cancel{\mathrm{\infty }}wrong$$

Commented by Ml last updated on 05/Jan/23

$$\mathrm{solution}$$

Answered by Frix last updated on 06/Jan/23

$$\mathrm{let}-\ln \frac{21}{10}=b$$$$-1<b<0\wedge x>0\Rightarrow$$$$b=\mid b\mid {\mathrm{e}}^{\mathrm{i}\pi }\Rightarrow b^{2x}=\mid b{\mid }^{2x}{\mathrm{e}}^{2\pi x\mathrm{i}}$$$$\mid b\mid <1\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\mid b{\mid }^{2x}=0$$$${\mathrm{e}}^{2\pi x\mathrm{i}}=\cos 2\pi x+\mathrm{i}\sin 2\pi x$$$$-1⩽\cos 2\pi x⩽1\wedge -1⩽\sin 2\pi x⩽1\Rightarrow$$$$\lim \mid b{\mid }^{2x}{\mathrm{e}}^{2\pi x\mathrm{i}}=0+0\mathrm{i}=0$$

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