Limit of $(1-2^{-x})^x$ I am observing that $(1-2^{-x})^x \to 1$ as $x \to \infty$, but am having trouble proving this. Why does the $-x$ "beat" the $x$?
I thought of maybe considering that $$1-(1-2^{-n})^n = 2^{-n}(1+2^{-n}+2^{-2n}+\cdots+2^{-n(n-1)})\le2^{-n} \frac{1}{1-2^{-n}} \to 0,$$
as $n \to \infty$ and then noting that $(1-2^{-x})^x$ is continuous to go from integers $n$ to any real $x$.
Is there a more elegant solution, or any better intuition? Moreover, it seems that $(1-2^{-ax})^{bx}\to 1$ holds as well for any $a,b>0$.
 A: \begin{align*}
\lim_{x\to\infty} (1-2^{-x})^x&=\lim_{x\to\infty} e^{\ln(1-2^{-x})^x}\\
&=\lim_{x\to\infty} e^{x\ln(1-2^{-x})}\\
&=e^{\lim_{x\to\infty} x\ln(1-2^{-x})}\\ 
\end{align*}
now lets just focus on the exponent... 
\begin{align*}
 \lim_{x\to\infty} x\ln(1-2^{-x})&=\lim_{x\to\infty} \frac{\ln(1-2^{-x})}{\frac{1}{x}} \to \frac{0}{0} \tag{then by LHopital..}\\ 
&=\lim_{x\to\infty} \frac{\frac{1}{1-2^{-x}}(1-2^{-x})^\prime}{\frac{-1}{x^2}}    \\ 
&=\lim_{x\to\infty} \frac{\frac{2^{-x}\ln 2}{1-2^{-x}}}{\frac{-1}{x^2}}    \\ 
&=\lim_{x\to\infty} \frac{ x^2\ln 2}{1-2^x}   \to 0\tag{after lHopital$^2$}  \\ 
\end{align*}
so the limit is $$e^0=1$$
A: Look at the logarithm:
$$
\log(1-2^{-x})^x=x\log(1-2^{-x})\sim x\cdot(-2^{-x})\rightarrow 0.
$$
So
$$
(1-2^{-x})^x \rightarrow 1.
$$
A: $(1-2^{-ax})^{bx}=e^{bx \log (1-2^{-ax})}$
We use L'Hospital's rule to calculate the limit:
$\lim_{x\to \infty}x \log (1-2^{-ax})=\lim_{x\to \infty}\frac{\log (1-2^{-ax})}{\frac{1}{x}}=\lim_{x\to \infty}\frac{-2^{-ax}\log 2 (-a)}{(1-2^{-ax})}-x^2=\lim_{x\to \infty}-a\log2 \frac{x^2}{2^{ax}-1}\to 0$ 
Hence the limit is 1 $\forall a,b>0$.
A: $(1-2^{-x})^x=\exp(x\cdot\ln(1-2^{-x}))$, so it is enough to prove that
$$
\lim_{x\to\infty}x\cdot\ln(1-2^{-x})=0.
$$
But, from the Maclaurin expansion, for big $x$, $\ln(1-2^{-x})\approx-2^{-x}$ and
$$
\lim_{x\to\infty}\frac{x}{2^x}=0.
$$
A: It is enough to show that, for $0 \le y \lt 1 \lt x$ we have $$1 \gt (1-y)^x \gt 1-yx$$ Because then, for $y=2^{-x}$ or any $y=y(x)$ with $yx \rightarrow 0,$ we have $$\lim_{x \rightarrow \infty}(1-y)^x=1.$$
Note: I think that the short proof below can be adjusted to show that, provided $yx \lt 1 \text{ and } 2 \le x,$  $$1-yx+y^2\frac{x(x-1)}{2} \gt (1-y)^x \gt 1-yx$$ which means that, for your problem, the estimate is quite accurate.
So fix $y$ and note that for $x=1$ $$(1-y)^x = (1-yx).$$ Then derivatives show that, as $x$ increases, the lefthand side decreases less rapidly than the right: $$-x(1-y)^{x-1} \gt -x.$$
That is actually a good estimate for your question. For $x=9$ and $y=2^{-9}=\frac{1}{512}$ we have $$ \begin{eqnarray*} 1-yx&=&0.982421875\\ (1-y)^x&=&0.98255858\cdots\\
1-yx+y^2\frac{x(x-1)}2&=&0.9825592\cdots \end{eqnarray*}$$
