An approximation for $\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y$ For large $x$ it is common to use the approximation $$\left(1-{1\over x}\right)^{x} \approx e^{-1}$$ leading to $$\left(1-{1\over x}\right)^{y} \approx e^{-y/x}$$ and so $$\left(1-{2\over x}\right)^y\approx e^{-2y/x}\approx \left(1-{1\over x}\right)^{2y}$$
But is there a simple approximation for the difference
$$\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y \approx \,?$$
When $x=y$ the difference is about $\frac{e^{-2}}{y}$ but the pattern seems less obvious when they are not equal.
 A: We have
$$\left( 1+ \frac{a}{x}\right)^x = e^a \left( 1 - \frac{a^2}{2x} +O(x^{-2}) \right)$$
Using this, assuming $y/x = a$ (constant) we get
$$\left(1-{1\over x}\right)^{2y} - \left(1-{2\over x}\right)^y \approx e^{-2 a} \frac{a}{x}=e^{-2 y/x} \frac{y}{x^2} $$
A: In the spirit as @leonbloy, we can make the problem more general considering
$$z=\left(1-\frac{k}{x}\right)^{b y}-\left(1-\frac{b k}{x}\right)^y$$ with $y=a x$. Using the same Taylor expansion, this would give
$$z=\frac{a (b-1) b k^2 e^{-a b k}}{2 x}\left(1-\frac{(b+1) k (3 a b k-8)}{12 x}+O\left(\frac{1}{x^2}\right)\right)$$ and, back to $y$, this becomes
$$z=\frac{b(b-1)k^2}{2}  \,\frac{y }{x^2}\exp\left({-b k\frac{ y}{x}}\right)+\cdots$$
Edit
As @Henry pointed out in a comment, it would be simpler to let $\frac k x = \frac 1 X$ and write
$$z=\left(1-\frac{1}{X}\right)^{b y}-\left(1-\frac{b }{X}\right)^y$$ with $y=a X$. For large $X$
$$z=\frac{a (b-1) b e^{-a b}}{2 X}\left(1-\frac{\left(b+1\right) (3 a b-8)}{12 X}+O\left(\frac{1}{X^2}\right) \right)$$ and, back to $y$,
$$z=\frac{b(b-1)}{2}  \,\frac{y }{X^2}\exp\left({-b \frac{ y}{X}}\right)+\cdots$$
