Show that $ \lim_{n \rightarrow \infty} (1 - b^n)^n= 1$ for $b \in [0, 1)$ This is a self study problem, I 'm trying to see how to go about proving this. Frankly, I don't even know where to start. 
 A: Using Bernoulli's inequality,
$$1 - nb^n \leqslant (1-b^n)^n \leqslant 1$$
By the squeeze theorem $(1-b^n)^n \to 1$ as $n \to \infty$ since $nb^n \to 0$ for $0\leqslant b< 1$.
Note that since $b < 1$, we have $b = 1/(1+a)$ for $a> 0$ and as $n \to \infty$,
$$nb^n = \frac{n}{(1+a)^n} < \frac{n}{\frac{n(n-1)}{2}a^2} = \frac{2}{(n-1)a^2 } \to 0$$ 
A: Do you know that $nb^{n} \to 0$ and $(1-\frac t n)^{n} \to e^{-t}$?. If yes then take $\epsilon >0$ and note that $b^{n} <\frac {\epsilon} n$ for $n$ sufiiciently large. This gives $(1-b^{n})^{n} >(1-\frac {\epsilon} n)^{n} \to e^{-\epsilon}$. Since $\epsilon $ is arbitatry we get $\lim \inf (1-b^{n})^{n} \geq 1$. Since $(1-b^{n})^{n} \leq 1$ for all $n$ we are done. 
A: $$\left(1-b^n\right)^n = e^{n \ln\left(1-b^n\right)} = e^{nb^n \frac{\ln\left(1-b^n\right)}{b^n}} $$
Since $b < 1$, $b^n \to 0$, $nb^n\to 0$ and $$\lim_{n\to \infty} \frac{\ln\left(1-b^n\right)}{b^n} = -1$$ so $$\lim_{n\to \infty}\left(1-b^n\right)^n = e^{0 \times (-1)} = 1$$
A: Define $L= \lim_{n \to \infty}n \ln (1-b^n)$.  Then our answer is $e^L$.
$$L=\lim_{n \to \infty} n \ln (1-b^n) = \lim_{n \to \infty} \frac{\ln(1-b^n)}{\frac{1}{n}}.$$
Because $b \in [0, 1)$, the latter limit is of the form $0/0$ so we can use L'Hopital's Rule:
$$L=\lim_{n \to \infty} \frac{\ln(1-b^n)}{\frac{1}{n}}= \lim_{n \to \infty} \frac{\frac{b^n  \ln b}{1-b^n}}{\frac{1}{n^2}}= \lim_{n \to \infty} \frac{n^2~b^n \ln b}{1-b^n}=0 \Rightarrow e^L=1.$$
A: The lemma by Thomas Andrews can be used here:

Lemma: If $n(a_n-1)\to 0$ then $a_n^n\to 1$.

For the current problem just take $a_n=1-b^n$ and show that $nb^n\to 0$ (you can use ratio test for this).
