# Thread: Limit -- Variation on a problem

1. ## Limit -- Variation on a problem

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

2. ## Re: Limit -- Variation on a problem

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.

3. ## Re: Limit -- Variation on a problem Originally Posted by HallsofIvy 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?

4. ## Re: Limit -- Variation on a problem

Not actually a typo- I just misread the problem!

5. ## Re: Limit -- Variation on a problem

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