What is the limit of $2^n\cdot(1-\exp(-a/2^n))$ for $a>0$

Question

I am trying to apply the Poisson limit theorem for a random variable that is

$\operatorname{Bin}(2^n,(1-\exp(-a/2^n))$- distributed. This means I have to calculate $\lim\limits_{n \rightarrow \infty}$ $2^n(1-\exp(-a/2^n))$ for some $a>0$ but have no idea how to do this.

The answer should be $\lim\limits_{n \rightarrow \infty}$ $2^n(1-\exp(-a/2^n))=a$ but even knowing this I can't come up with a proof.

Answer 1

Hint: Let $2^n=\dfrac1t$ and use L'Hospital rule $$\lim_{n\to\infty}2^n(1-\exp(-a/2^n))=\lim_{t\to0}\dfrac{1-e^{-at}}{t}=a$$

Answer 2

Recall that as $x=\frac a {2^n}\to 0$

$$e^x=1+x+o(x)$$

therefore

$$2^n\left(1-e^{-\frac a {2^n}}\right)=\frac a x\left(1-e^{-x}\right)=\frac a x\left(1-1+x+o(x)\right)=a+o(1)\to a$$

or by standard limit as $t \to 0 \quad \frac{e^t-1}{t}\to 1$

$$2^n\left(1-e^{-\frac a {2^n}}\right)=\frac a x\left(1-e^{-x}\right)=a\frac{1-e^{-x}}{x}=a\frac1{e^x}\frac{e^x-1}{x}\to a\cdot 1\cdot 1=a$$

Answer 3

Without invoking L'Hospital, it is well-known that $$1-\mathrm e^u\sim_0 -u,\quad\text{hence}\quad1-\mathrm e^{-\tfrac a{2^n}}\sim_{n\to\infty}\frac a{2^n},$$ so that in the end $$2^n\Bigl(1-\mathrm e^{-\tfrac a{2^n}}\Bigr)\sim_{n\to\infty}2^n\frac a{2^n}=a.$$