How would one calculate $\lim_{n\to\infty}n((1+\frac{1}{n})^n-e)$?

Question

What I have thought about this is:

we may use L'Hopstal's rule to calculate $\lim_{n\to\infty}\frac{((1+\frac{1}{n})^n-e)}{\frac{1}{n}}$, both the numerator and denominator goes to 0 as n goes to infinity. But calculating the derivative of $(1+1/n)^n$ seems to be very complicated.
Using Taylor series to calculate the dominant terms of $(1+1/n)^n$, but I'm not really sure if it makes sense to let $"n=\infty"$. Equivalently maybe we can expand $(1+x)^{1/x}$ at $x=0$, but it's not defined.

Maybe I wasn't on the right track. Even if the solution uses a different approach, I would still love to know how to expand $(1+x)^{1/x}$. Thanks for any suggestions.

Hi Thanks that solves my problem. Still I'm curious how we can expand $(1+x)^{1/x}$ EDIT: Nvm seems like we can do a $e^{\log(\dot)}$ and use taylor series twice, thanks! — user3799934, Sep 16 '17 at 21:05
We may use Salahamam_ Fatima's hint, or you may consider finding the Maclaurin expansion of $$f(x)=\begin{cases}(1+x)^{1/x},&x\ne0\e,&x=0\end{cases}$$which should give you the same conclusion as the aforementioned answer. — Simply Beautiful Art, Sep 16 '17 at 21:07

hamam_Abdallah · Accepted Answer · 2017-09-16T20:58:28.617

2

hint

$$f (x)=(1+x)^\frac 1x=e^{\frac {1}{x}\ln (1+x)} $$

$$\ln (1+x)=x-\frac {x^2}{2}(1+\epsilon (x)) $$

$$f (x)=e.e^{-\frac {x}{2}(1+\epsilon (x))} $$

$$=e\Bigl (1-\frac {x}{2}(1+\epsilon (x)\Bigr)$$

The limit will be $$-\frac {e}{2} $$

edited Sep 16 '17 at 20:58

answered Sep 16 '17 at 20:52

hamam_Abdallah

62,951

@Isham What didn't you understand. – hamam_Abdallah Sep 16 '17 at 21:08
@Fatima Well I think I should read it again to understand clearly what you wrote... – user577215664 Sep 16 '17 at 21:10

How would one calculate $\lim_{n\to\infty}n((1+\frac{1}{n})^n-e)$?

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