Why a number which is in (-1,0) raised to infinity is 0?

Question

I hope it's not a duplicate but I've been searching about this problem for some time on this site and I couldn't find anything. My problem is why a number $\in(-1,0)$ raised to $\infty$ is $0$. For example let's take $$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n$$ Which is equivalent to $$\left(\frac{-1}{2}\right)^\infty=(-1)^\infty\left(\frac{1}{2}\right)^\infty=0(-1)^\infty$$ But if a sequence converges all its subsequences converge to the same limit. And $(-1)^{2n}$ is a subsequence of $(-1)^n$ that converges to $1$ when $(-1)^{2n + 1}$ is a subsequence of $(-1)^n$ that converges to $-1$. So $(-1)^\infty$ does not exist. It remains that$$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n=0(DNE)$$

Answer 1

Hint:

Indeed $(-1)^\infty$ does not exists. To show that $(-1/2)^n$ goes to $0$, an idea is to show that: $$ \lim_{n \to \infty}\left|\left(\frac{-1}{2} \right)^n-0 \right|=0$$

Answer 2

Your assertion that $(\frac{-1}{2})^\infty=(-1)^\infty(\frac{1}{2})^\infty$ is wrong because the first limit on the right side does not exist. The limit of a product is only equal to the product of limits when all the limits exist.

Answer 3

If $\lim a_n = L$ exist and $a_n = b_n*c_n$ it does not follow that $\lim b_n$ exists and, indeed, we can ALWAYS find counter examples. (Ex: $\lim \frac 1{2^n} = 0$ but $\frac 1{2^n} = 2^n\frac 1{2^{2n}}$ but $\lim 2^n$ is not finite.)

So $\lim (\frac {-1}2)^n = (-1)^{n}*(\frac {1}2)^n$ but $\lim (-1)^n$ does not exist is utterly irrelevant.

===

What is true is that if $a_n = b_n*c_n$ and $\lim b_n$ and $\lim c_n$ both exist, then $\lim a_n = \lim b_n*\lim c_n$.

And if $a_n = b_n*c_n$ and $\lim a_n$ exists and $\lim c_n$ exists AND $\lim c_n \ne 0$ then $\lim c_n = \frac {\lim a_n}{\lim b_n}$.

Neither of those are the case here.