Taking Limit to infinity when variable is exponent? Example. I am testing $-1/4$ and $1/4$ for
$$\lim_{n \to  \infty} \frac{(-1)^n 4^n x^n}{\sqrt{n}}$$
What happens in the numerator that it makes it equal 1?
Like what happens to each term that it everything ends up as 1? 
Any help would be greatly super appreciated! Thanks
 A: For $x = -1/4$, you have
$$\lim_{n \to \infty} \frac{(-1)^n 4^n (-1/4)^n}{\sqrt{n}}.$$
Since you have a bunch of terms all raised to the same power in the numerator, we can simplify the equation a bit by pulling these terms together inside one pair of parenthesis:
$$\lim_{n \to \infty} \frac{\big[(-1)(4)(-1/4)\big]^n}{\sqrt{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n}}.$$
I skipped a few steps I thought were obvious. I encourage you try working everything out for $x = 1/4.$
A: 
What do I do to these undefined terms when they go to infinity. Am I missing something about limits?

The implication of using the variable $n$, especially given the expression you're taking the limit of, is that the variable is meant to range only over integers, rather than over all real numbers.
And similarly, in context, exponentiation is clearly meant to be the discrete version: so for $a>0$, rather than $(-a)^n$ being undefined because it has a negative base, instead it is equal to
$$ (-a)^n = \begin{cases} -(a^n) & n \text{ odd} 
\\ a^n & n \text{ even} \end{cases} $$
A: Observe that
$$
(-4)^n\times \left(-\frac14\right)^n=(-4)^n\times \left(\frac1{-4}\right)^n=(-4)^n\times \frac1{\left(-4\right)^n}=1.
$$
