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

lim_(x→∞)  (−ln((21)/(10)))^(2x) =?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(−\mathrm{ln}\frac{\mathrm{21}}{\mathrm{10}}\right)^{\mathrm{2x}} =? \\ $$

Answered by SEKRET last updated on 05/Jan/23

0

$$\mathrm{0} \\ $$$$ \\ $$

Commented by Ml last updated on 05/Jan/23

solution

$$\mathrm{solution} \\ $$

Answered by JDamian last updated on 05/Jan/23

∞ wrong

$$\cancel{\infty}\:{wrong} \\ $$

Commented by Ml last updated on 05/Jan/23

solution

$$\mathrm{solution} \\ $$

Answered by Frix last updated on 06/Jan/23

let −ln ((21)/(10)) =b  −1<b<0∧x>0 ⇒  b=∣b∣e^(iπ)  ⇒ b^(2x) =∣b∣^(2x) e^(2πxi)   ∣b∣<1 ⇒ lim_(x→∞)  ∣b∣^(2x)  =0  e^(2πxi) =cos 2πx +i sin 2πx  −1≤cos 2πx ≤1 ∧ −1≤sin 2πx ≤1 ⇒  lim ∣b∣^(2x) e^(2πxi)  =0+0i=0

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

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