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

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?

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

Solution 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.

Solution 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.

Solution 3:

By the L'Hôpital's rule we obtain: $$\lim_{n\rightarrow+\infty}n\left(1-\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}\right)=\lim_{x\rightarrow0^+}\frac{e-(1+x)^{\frac{1}{x}}}{ex}=-\frac{1}{e}\lim_{x\rightarrow0^+}\left(e^{\frac{\ln(1+x)}{x}}\right)'=$$ $$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\left(\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}\right)=$$ $$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\lim_{x\rightarrow0^+}\frac{x-(1+x)\ln(1+x)}{x^2+x^3}=$$ $$=-\lim_{x\rightarrow0^+}\frac{1-\ln(1+x)-1}{2x+3x^2}=\lim_{x\rightarrow0^+}\frac{\ln(1+x)}{x}\lim_{x\rightarrow0^+}\frac{1}{2+3x}=\frac{1}{2}.$$