Is $x^n$ for some $0A negligible function is a function that goes faster towards zero than the inverse of any polynomial (when the variable goes to infinity). For example $g(n) = \frac{1}{2^n}$ is a negligible function.
Is it in general true that $f(n) = x^n$ for some $0<x<1$ a negligible function? Note that x is a constant and n is the variable here.
If not, is $h(n) = \left(\frac{3}{4}\right)^n$ a negligible function? How would I show that?
 A: Rewrite the function as 
$$f(n) = e^{-n \log{(1/x)}}$$
As a function of $n$, because $\log{(1/x)} \gt 0$, $f$ decreases faster than any inverse polynomial as $n \to \infty$. Thus, according to your definition, $f(n)$ is "negligible."
A: It is negligible. To show that $f\left(n\right)$ goes to zero faster than the inverse of any polynomial in $n$, just show that the ratio $f\left(n\right)/q\left(n\right)$ goes to zero, where $q$ is the inverse of a polynomial:
$$\left|\frac {f\left(n\right)} {q\left(n\right)}\right| = {x^n}\cdot\left| {a_d n^d + a_{d-1} n^{d-1} + \cdots}\right| =
\left|a_d\right| \cdot {x^n}{n^d}\cdot \left|{1+a_{d-1}n^{-1}/a_d+a_{d-2}n^{-2}/a_d+\cdots}\right|
$$
If you can show that $x^n n^d$ goes to zero as $n\to\infty$, with $d$ fixed, then you have shown that $f\left(n\right)$ is negligible in your definition. You also need to show that the quantity at the right goes to $1$, but that should be easy.
A: Standard argument: Let $d=\deg p$ be the degree of the polynomial, set $|x|=1/(1+a)$, then for $n>d$
$$
|p(n)x^n|\le \frac{|p(n)|}{1+\binom{n}{d+1}a^{d+1}}\sim\frac1n
$$
where the denominator has one degree more than the numerator and thus the bound converges to zero.
