ПОМОГИТЕ ПОЖАЛУЙСТА . Created by Ternov21. matematika-ru

ПОМОГИТЕ ПОЖАЛУЙСТА ...

Ternov21 @Ternov21

August 2022 1 45 Report

ПОМОГИТЕ ПОЖАЛУЙСТА

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NNNLLL54

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$\boxed{\ \ \lim\limits_{\alpha (x) \to \infty}\, \left(1+\dfrac{1}{\alpha (x)}\right)^{\alpha (x)}=e\ \ ,\ \ \ \ \ \lim\limits_{\alpha (x) \to 0}\, \Big(1+\alpha (x)\Big)^{\frac{1}{\alpha (x)}}=e\ \ }$

$1)\ \ \lim\limits_{x \to \infty}\, \Big(1+\dfrac{5}{3x}\Big)^{2x}=\lim\limits_{x \to \infty}\, \Big(1+\dfrac{5}{3x}\Big)^{\frac{3x}{5}\cdot \frac{5}{3x}\cdot 2x}=\lim\limits_{x \to \infty}\, \Big(\underbrace{\Big(1+\dfrac{5}{3x}\Big)^{\frac{3x}{5}}}_{\to \ e}\right)\Big)^{\frac{10}{3}}=\\\\\\=\lim\limits_{x \to \infty}\, e^{\frac{10}{3}}=\sqrt[3]{e^{10}}$

$2)\ \ \lim\limits_{x \to \infty}\, \Big(1+\dfrac{2}{3x}\Big)^{x}=\lim\limits_{x \to \infty}\, \Big(1+\dfrac{2}{3x}\Big)^{\frac{3x}{2}\cdot \frac{2}{3x}\cdot x}=\lim\limits_{x \to \infty}\, \Big(\underbrace{\Big(1+\dfrac{5}{3x}\Big)^{\frac{3x}{2}}}_{\to \ e}\right)\Big)^{\frac{2}{3}}=\\\\\\=\lim\limits_{x \to \infty}\, e^{\frac{2}{3}}=\sqrt[3]{e^{2}}$

$3)\ \ \lim\limits_{x \to 0}\, \Big(1+2x\Big)^{\frac{5}{x}}=\lim\limits_{x \to \infty}\, \Big(1+2x\Big)^{\frac{1}{2x}\cdot 2x\cdot \frac{5}{x}}=\lim\limits_{x \to \infty}\, \Big(\underbrace{\Big(1+2x\Big)^{\frac{1}{2x}}}_{\to \ e}\right)\Big)^{10}=\\\\\\=\lim\limits_{x \to \infty}\, e^{10}=e^{10}$

$4)\ \ \lim\limits_{x \to \infty}\, \Big(1+\dfrac{1}{x}\Big)^{5x}=\lim\limits_{x \to \infty}\, \Big(1+\dfrac{1}{x}\Big)^{x\cdot 5}=\lim\limits_{x \to \infty}\, \Big(\underbrace{\Big(1+\dfrac{1}{x}\Big)^{x}}_{\to \ e}\right)\Big)^{5}=\lim\limits_{x \to \infty}\, e^{5}=e^{5}$

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