Find $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$

Question

$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$

I let, $x = \frac{1}{n}$, then as

$\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$

L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$

Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.

Answer 1

By Taylor's theorem with Peano's remainder, we have \begin{align*} & \left(1 + \frac{1}{n}\right)^n = \exp\left(n\log\left(1 + \frac{1}{n}\right)\right)\\ = & \exp\left(n\left(\frac{1}{n} - \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)\right) \right)\\ = & \exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right) \end{align*}

Now expand $\exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right)$ at $1$: $$\exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right) = e + e \times \left(-\frac{1}{2n} + o\left(\frac{1}{n}\right)\right) + o\left(\frac{1}{n}\right)$$ Therefore, $$n\left[\left(1 + \frac{1}{n}\right)^n - e\right] = -\frac{1}{2}e + o(1)\to -\frac{1}{2}e$$ as $n \to \infty$.

Answer 2

note that :in line 4 you missed something ,that is $$y=(1+x)^{\frac{1}{x}} ,y' =?\\\ln y=\ln(1+x)^{\frac{1}{x}} \\\ln y=\frac{1}{x} \ln(1+x)\\\frac{y'}{y}=\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\\y'= (1+x)^{\frac{1}{x}}\cdot \bigg(\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\bigg)$$