William Collins

2022-01-04

Compute the limit of the sequences:
$\left(a\right){\left(1-\frac{3}{n}\right)}^{2n}$
$\left(b\right){3}^{\frac{1}{n}}$

Shawn Kim

(a)
$\left(a\right){\left(1-\frac{3}{n}\right)}^{2n}$
$\underset{n\to d}{lim}{\left(1-\frac{3}{n}\right)}^{2n}$
$=\underset{n\to d}{lim}{e}^{\mathrm{ln}\left[{\left(1-\frac{3}{n}\right)}^{2n}\right]}$
$=\underset{n\to d}{lim}{e}^{2n×\mathrm{ln}\left(1-\frac{3}{n}\right\}}$
$={e}^{\underset{n\to d}{lim}2n×\mathrm{ln}\left(1-\frac{3}{n}\right)}$
$={e}^{2\underset{n\to d}{lim}n×\mathrm{ln}\left(1-\frac{3}{n}\right)}$
$={e}^{\underset{n\to d}{lim}}\frac{\mathrm{ln}\left(1-\frac{3}{n}\right)}{\frac{1}{n}}$
${e}^{\underset{n\to d}{lim}}\left(\frac{\frac{-1}{1-\frac{3}{n}}x}{\frac{1}{{n}^{2}}}\right)$
${e}^{2}x\left\{\underset{n\to d}{lim}\right\}\frac{-3}{1-\frac{3}{n}}$
${e}^{2x\left(-3\right)\underset{n\to d}{lim}}\frac{1}{1-\frac{3}{n}}$
${e}^{-6x}\frac{1}{1-\frac{3}{n}}={e}^{-6x\frac{1}{1-0}}$
${e}^{-6×1}={e}^{-6}$
(b)
${3}^{\frac{1}{n}}$
$\underset{n\to d}{lim}{3}^{\frac{1}{n}}$
${3}^{0}=1$

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