<s>[INST] Find $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$ $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$
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. 
 [/INST] 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 $n \to \infty$.
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