Destinee Hensley

2022-04-04

Facing difficulty finding limit
$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x}{x-1}\right)}^{2x+1}$

membatas0v2v

If you know that
$\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{a}{x}\right)}^{x}={e}^{a}$
so that
$\underset{x\to \mathrm{\infty }}{lim}{\left(1-\frac{1}{x}\right)}^{x}={e}^{-1}$
then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,
${\left(\frac{x}{x-1}\right)}^{2x+1}={\left({\left(\frac{x}{x-1}\right)}^{x}\right)}^{2}\left(\frac{x}{x-1}\right)$
$={\left(\frac{1}{{\left(\frac{x-1}{x}\right)}^{x}}\right)}^{2}\left(\frac{x}{x-1}\right)$
$={\left(\frac{1}{{\left(1-\frac{1}{x}\right)}^{x}}\right)}^{2}\left(\frac{x}{x-1}\right)$

clarkchica44klt

$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x}{x-1}\right)}^{2x+1}=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x-1+1}{x-1}\right)}^{2x+1}$
$=\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x-1}\right)}^{2x+1}=\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x-1}\right)}^{\left(x-1\right)\cdot \frac{2x+1}{x-1}}$
$=\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x-1}\right)}^{\left(x-1\right)\cdot \frac{2x+1}{x-1}}$
$={e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{2x+1}{x-1}}={e}^{2}$

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