Limits on successions: Indeterminate Form of ${1^n}$ and ${-1^n}$ I am currently studying limits of successions (therefore n can only approach positive infinity).
I'm reading my textbook and I can't understand why $1^n$ is an indeterminate form as 1 to the power of any real number (except for zero which I understand could be problematic) can only be 1.
Moreover, this also appeared in one of my textbook problems:
$\lim_n{\frac{n+(-1)^n}{n-(-1)^n}} = 1$
Although it can be solved as $\frac{n+(-1)^n}{n-(-1)^n} = \frac{1+\frac{(-1)^n}{n}}{1-\frac{(-1)^n}{n}} \implies \lim_n{\frac{1+0}{1-0}} = 1$ since a limited succession over infinity approaches one, I was wondering on how one could get out of the indeterminate form as my textbook is placing a certain emphasis on being able to transform successions to avoid indeterminate forms.
Would it be correct (in this specific case) to multiply by $\frac{-2^n}{-2^n}$ so that $\frac{n+(-1)^n}{n-(-1)^n} = \frac{n(-2^n)+(2)^n}{n(-2^n)-(2)^n} = \frac{\frac{n(-2^n)}{2^n}+1}{\frac{n(-2^n)}{{2^n}}-1} = \frac{-n+1}{-n-1} = 1$?
Or would it be more correct to calculate separately -1 and 1 (which are the values $-1^n$ can be)?
CORRECTED FORMULAS (The ones typed wrongly)
$
\frac{n+(-1)^n}{n-(-1)^n} =
\frac{(n+(-1)^n)*((-2)^n)} {(n-(-1)^n)*((-2)^n)} = 
\frac{n*((-2)^n)+(2)^n}{n*((-2)^n)-(2)^n}$ which is heavily different from what I typed originally and still returns to a $(-1)^n$ situation.
 A: No, that is not correct, since, if it was, you would have $\dfrac{n+(-1)^n}{n-(-1)^n}=1$ for every $n\in\Bbb N$. In other words, you would always have $n+(-1)^n=n-(-1)^n$, which is actually never true.
Your error lies in the computation of $\bigl(n\pm(-1)^n\bigr)\times(-2^n)$. What you did was to multiply $n$ by $-2^n$ and to multiply $(-1)^n$ by $(-2)^n$. But $(-2)^n=-2^n$ when and only when $n$ is odd.
A: Your first idea was the right one, don't change anything, except: don't write "$\lim_n\frac{1+0}{1-0}$". Simply
$$\lim_n\frac{1+\frac{(-1)^n}{n}}{1-\frac{(-1)^n}{n}}=\frac{1+0}{1-0}= 1.$$
The first equality is due to the fact that $\lim_n\frac{(-1)^n}n=0$ because the product of a (not necessarily converging) bounded sequence by a sequence which tends to $0$ also tends to $0$ (here: $(-1)^n\cdot\frac1n\to0$).
Btw, it is not $1^n$ and $(-1)^n$ which are indeterminate forms. They are sequences, the first one is constant, the second one has no limit. The true indeterminate forms are $1^{\pm\infty}.$ It means that if you only know $x_n\to1$ and $y_n\to+\infty$ or $-\infty,$ you can say nothing about $\lim {x_n}^{y_n}.$
