Limit -- Variation on a problem

Dec 2016
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Earth
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\(\displaystyle \displaystyle\lim_{n \to \infty}\bigg[n*e \ - \ n*(1 + \tfrac{1}{n})^n\bigg]\)
 
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HallsofIvy

MHF Helper
Apr 2005
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First note that the discrete limit, \(\displaystyle \lim_{n\to \infty} ne- n\left(n+ \frac{1}{n}\right)^n\) is the same as the continuous limit \(\displaystyle \lim_{x\to\infty} xe- x\left(x+ \frac{1}{x}\right)^x\) as long as the latter limit exists. To do that, write it as \(\displaystyle \lim_{x\to\infty}\frac{2- \left(x+ \frac{1}{x}\right)^x}{\frac{1}{x}}\), where both numerator go to 0, and use L'Hopital's rule.
 

Plato

MHF Helper
Aug 2006
22,469
8,640
First note that the discrete limit, \(\displaystyle \lim_{n\to \infty} ne- n\left(\color{red}n+ \frac{1}{n}\right)^n\) is the same as the continuous limit \(\displaystyle \large\lim_{x\to\infty} xe- x\left(\color{red}x+ \frac{1}{x}\right)^x\) as long as the latter limit exists. To do that, write it as \(\displaystyle \large\lim_{x\to\infty}\frac{\color{red}2- \left(\color{red}x+ \frac{1}{x}\right)^x}{\frac{1}{x}}\), where both numerator go to 0, and use L'Hopital's rule.
Prof. Ivey, are there typo in the above?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Not actually a typo- I just misread the problem!
 
Jun 2013
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Lebanon
we can use L'Hôpital's rule a couple of times or write

\(\displaystyle e-\left(1+\frac{1}{n}\right)^n=\sum _{k=2}^n \left(\frac{1}{k!}-\left(
\begin{array}{c}
n \\
k
\end{array}
\right)\frac{1}{n^k}\right)+\sum _{k=n+1}^{\infty } \frac{1}{k!}\)

using the series for $e$ and the binomial theorem to expand $\left(1+\frac{1}{n}\right)^n$

do some algebra then multiply by $n$ and take the limit
 
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