$\lim_{x \to \infty } {1.0000000000001^x}/{x^{100}} = 0$

Question

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?

Answer 1

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 $\frac{\left(1+10^{-13}\right)^x}{x^{100}}\gt1$.

Answer 2

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.