Destinee Hensley

2022-04-04

Facing difficulty finding limit

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

membatas0v2v

Beginner2022-04-05Added 19 answers

If you know that

$\underset{x\to \mathrm{\infty}}{lim}{(1+\frac{a}{x})}^{x}={e}^{a}$

so that

$\underset{x\to \mathrm{\infty}}{lim}{(1-\frac{1}{x})}^{x}={e}^{-1}$

then you can try to rewrite your limit into something involving this limit.

So try rewriting it; perhaps as a product,

${\left(\frac{x}{x-1}\right)}^{2x+1}={\left({\left(\frac{x}{x-1}\right)}^{x}\right)}^{2}\left(\frac{x}{x-1}\right)$

$={\left(\frac{1}{{\left(\frac{x-1}{x}\right)}^{x}}\right)}^{2}\left(\frac{x}{x-1}\right)$

$={\left(\frac{1}{{(1-\frac{1}{x})}^{x}}\right)}^{2}\left(\frac{x}{x-1}\right)$

so that

then you can try to rewrite your limit into something involving this limit.

So try rewriting it; perhaps as a product,

clarkchica44klt

Beginner2022-04-06Added 17 answers

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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