How do I find this limit: $\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right]$?

Question

I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?

$$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$

Answer 1

Just use the fact that $$\left(1+\frac1n\right)^n=e^{n\ln(1+\frac1n)}=e^{n\left(\frac1n-\frac{1}{2n^2}+o(\frac{1}{n^2})\right)}=e\cdot e^{-\frac{1}{2n}+o(\frac1n)}=e\left(1-\frac{1}{2n}+o\Bigl(\frac{1}{n^2}\Bigr)\right),$$and the result follows.

Answer 2

Hint:

$$\lim \limits_{n\rightarrow \infty} n-\frac{n}{e}\left(1 +\frac{1}{n}\right)^n = \lim \limits_{n\rightarrow \infty} \frac{1-\frac{1}{e}\left(1 +\frac{1}{n}\right)^n}{\frac{1}{n}},$$ and use l'Hopital rule.

Answer 3

Hint: try logarithms. $$f(n) - g(n) = \ln\frac{f(n)}{g(n)}$$