Approximation for $(1 - \frac{1}{x})^x$
We have: $$ \left(1 - \frac{1}{x}\right)^x = e^{x\log\left(1-\frac{1}{x}\right)} = e^{x\left(-\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right)} = e^{\left(-{1}-\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)}$$
by Taylor expansion of $\log(1-t)$ at $0$. Then by expanding the exponential : $$ e^{\left(-\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)} = 1-\frac{1}{2x}+o\left(\frac{1}{x}\right) + o\left(\frac{1}{x}\right) = 1+O(1/x) $$
So that plugging everything back: $$\left(1 - \frac{1}{x}\right)^x = e^{-1}(1+O(1/x)) = e^{-1}+O(1/x)$$