proof that $x/c^x$ goes to 0 Im trying to show that $\lim_{x\to\infty}\frac{x}{c^x}$ where $c>1$ is a constant goes to zero.  To show this i know i need to take some $\epsilon$ and find some $m$ such that $\forall x>m$, $|\frac{x}{c^x}|<\epsilon$.  However, im having trouble finding $m$. What is the usual way people go to find $m$?
 A: First let's do the case where $x \to \infty$ along the integers. At the very beginning of a rigorous analysis course (certainly well before we define exponential function, or logarithm, or derivative, or irrational powers) we may prove Bernoulli's inequality by induction on $n$:
$$
(1+\delta)^n \ge 1+\delta n,\qquad\text{for }\delta>0, n \in \mathbb N .
$$
From that, we get:  for $a>1$
$$
a^n \ge (a-1)n
$$
and deduce that
$$
a^n \to \infty\qquad\text{and also that }\qquad\frac{a^n}{n} \ge (a-1).
$$  Given any $c>1$, I can write it as $c=ab$, where $a>1, b>1$. (I can do this without knowing square roots exist).  Then
$$
\frac{c^n}{n} = a^n\;\cdot\;\frac{b^n}{n} \ge a^n (b-1)
$$
But $b-1 > 1$ and $a^n \to \infty$, so we conclude $\frac{c^n}{n} \to \infty$.
[This argument is valid in an archimedian ordered field.  In particular it is valid even if irrational powers and/or square roots do not exist in the field.]
added
What about the case where $x \to \infty$ in the reals?  Of course we need to know advanced things like: how to define $c^x$ and what its properties are.  But then we can proceed like this.
Given $x > 2$, there is $ n  \in \mathbb N$ with $n\le x < n+1 < 2n$.  Then
$$
c^x \ge c^n
\\
x < 2n
\\
\frac{c^x}{x} \ge \frac{c^n}{2n} = \frac{1}{2}\cdot\frac{c^n}{n} .
$$
Therefore $\frac{c^x}{x} \to \infty$ also.
A: Hint: Assume that $c\gt 1$.  Then we are looking for the limit of $\frac{x}{e^{x\ln c}}$.
By the power series expansion for $e^t$, we have for positive $t$ that $e^t\gt 1+t+\frac{t^2}{2}\gt \frac{t^2}{2}$. Now it should not be hard to find a suitable $m$.
