Showing $\lim_{n\to\infty} (1-x)^n = 0$ for $0<x<1$

Question

I believe this can probably be shown by applying the binomial theorem, but I am wondering if there is another (hopefully simpler) way

Answer 1

The prospect of using the binomial theorem was mentioned in the OP. Here, we present an approach that relies on the Bernoulli's Inequality, which is more elementary than the binomial theorem. To that end, we now proceed.

---

Note that for any $0<x<1$,

$$0<(1-x)^n<\frac{1}{(1+x)^n}\tag1$$

Using Berenoulli's inequality on the right-hand side of $(1)$ reveals

$$0<(1-x)^n<\frac{1}{1+nx}\tag2$$

Applying the squeeze theorem to $(2)$ yields the coveted limit

$$\lim_{n\to\infty}(1-x)^n=0$$

for $x\in(0,1)$.

Answer 2

Let $y=1-x$. Then $0 < y < 1$ and so $y^n \to 0$.

Answer 3

Note that $0<1-x<1$. Thus we just need to show that $a^n$ goes to zero as n goes to $\infty$. If you don’t know how to show this you can use the fact that there are rational numbers $q$ and $r$ such that $q < a < r$ and then use the Squeeze Theorem. It’s easy to show that for a rational number in $(0, 1)$, $r^n$ goes to zero in the limit.

Answer 4

You can also use the ratio test for sequences.

$$\frac{(1-x)^{n+1}}{(1-x)^n}=1-x<1$$ thus $(1-x)^n \to 0$

Or you can use geometric series:

$$\sum_{n=1}^{+\infty}(1-x)^n =\frac{1}{1-(1-x)}< \infty$$

If $\sum_{n}a_n$ converges then we know that $a_n \to 0$ so $(1-x)^n \to 0$