Prove that $\lim \limits _{n \to \infty} \left( n - \frac 1 {e^{\frac 1 n} - 1} \right) = \frac 1 2$

Question

Limit of reciprocal of nth root of e minus 1 is linear

I somehow got around to needing to know the value of $\frac 1 {e^{\frac 1 n} -1}$, and noticed that it seems very linear:

And that the difference from $n$ seems to go to $\frac 1 2$:

Thus I would like to see a proof of this fact, which I was unable to prove:
$$\lim \limits _{n \to \infty} \left( n - \frac 1 {e^{\frac 1 n} - 1} \right) = \frac 1 2$$

Answer 1

$$\lim_{N\to\infty} \left( N - \frac{1}{e^{\frac{1}{N}} -1} \right) = \lim_{x\to0}\left(\frac{1}{x} - \frac{1}{e^x -1}\right) = \lim_{x\to0} \frac{e^x - 1 -x}{x(e^x-1)}$$

Now apply L'Hospital's rule twice to get this being $$\lim_{x\to0} \frac{e^x}{2e^x +xe^x} = \lim_{x\to0} \frac{1}{2+x} = \frac{1}{2}.$$

Answer 2

$$e^{1/N}=1+\frac1N+\frac1{2N^2}+o(N^{-2})\\ e^{1/N}-1=\frac1N\left(1+\frac1{2N}+o(N^{-1})\right)\\ \frac1{e^{1/N}-1}=N\left(1+\frac1{2N}+o(N^{-1})\right)^{-1}\\ =N\left(1-\frac1{2N}+o(N^{-1})\right)$$

Answer 3

$$\lim \limits _{n \to \infty} \left( n - \frac 1 {e^{\frac 1 n} - 1} \right)=\lim \limits _{m \to 0} \left(\frac 1m - \frac 1 {e^m - 1} \right)=\lim \limits _{m \to 0}\frac{e^m-1-m}{m(e^m-1)}$$

Now from $e^m=1+m+\frac {m^2}{2!}+\frac{m^2}{3!}+o(m^3)$ we get $$\lim \limits _{m \to 0}\frac{e^m-1-m}{m(e^m-1)}=\lim \limits _{m \to 0}\frac{m^2(\frac12+\frac{m}{3!}+....)}{m^2(1+\frac m2+....)}=\lim \limits _{m \to 0}\frac{\frac12+\frac{m}{3!}+....}{1+\frac m2+....}=\frac12 $$