$\lim_{x \to \infty } {1.0000000000001^x}/{x^{100}} = 0$ I am confused as to how Wolfram alpha got the answer to be $0$, because I am fairly sure that the answer should be infinity, can somebody explain this to me?
 A: Since $e^x\ge1+x$,
$$
\begin{align}
\left(1+10^{-13}\right)^x
&=e^{x\log\left(1+10^{-13}\right)}\\[6pt]
&=e^{\left(\left.x\log\left(1+10^{-13}\right)\middle/200\right.\right)\cdot200}\\[6pt]
&\ge\left(1+\frac{x\log\left(1+10^{-13}\right)}{200}\right)^{200}\\
&\ge x^{200}\,\left(\frac{\log\left(1+10^{-13}\right)}{200}\right)^{200}
\end{align}
$$
Therefore,
$$
\frac{\left(1+10^{-13}\right)^x}{x^{100}}
\ge x^{100}\,\left(\frac{\log\left(1+10^{-13}\right)}{200}\right)^{200}\\
$$
and the limit is obviously infinite.
As mentioned in comments, the problem is most likely round-off error due to finite precision arithmetic.
As I mentioned in comments, it takes $x\gt38181117481549541$ to even get $\frac{\left(1+10^{-13}\right)^x}{x^{100}}\gt1$.
A: If
$0 < c < 1$
and
$a > 0$
then
$\begin{array}\\
f(x)
&=\ln(\dfrac{(1+c)^x}{x^a})\\
&=x\ln(1+c)-a\ln(x)\\
&\gt xc/2-a\ln(x)
\qquad\text{since } \ln(1+c) > c/2\\
&\gt xc/2-2a\sqrt{x}
\qquad\text{since } \ln(x) < 2\sqrt{x}\\
&=\sqrt{x}(\sqrt{x}c/2-2a)\\
\end{array}
$
so
$f(x) > 0$
for
$x > (4a/c)^2
$
and
$f(x) > \sqrt{x}$
for
$x > (2(2a+1)/c)^2
$.
In particular,
$f(x) \to \infty$.
These bounds, of course,
are much worse than 
the best bounds.
