How can I calculate $\lim_{ n \rightarrow \infty} (n^{1.000001}/\log n) / (1.000001)^n$

Question

I'm using the limit definition to show that $f(n) \in\mathcal{O}(g(n))$ or the reverse for the following functions;

$f(n) = n^{1.000001}/\log n$

$g(n) = (1.000001)^n$

I calculated the limit as $\approx 0$ in Wolfram Alpha.

$$\lim_{ n \rightarrow \infty} f(n)/g(n) \approx 0, $$ and

$$\lim_{ n \rightarrow \infty} g(n)/f(n) \approx 0.$$

This is not right. How can I calculate the limits?

Answer 1

Consider $1>a >0$, $f(x) = x^{1+a} /\log(x)$, $g(x)= (1+a)^x$. Then for every $x > 0$, $$ (1+a)^x = \mathrm e^{x \log(1+a)} = \sum_0^{+\infty} \frac {x^n \log(1+a)^n}{n!} > \frac {x^3 \log(1+a)^3}{6}, $$ then \begin{align*} 0 \leqslant \frac fg (x) &= \frac {x^{1+a}}{(1+a)^x \log(x)} \\&< \frac {6x^2}{x^3 \log(1+a)^3\log(x)} \\&<\frac 6{x \log(1+a)^3} \xrightarrow{x \to +\infty} 0, \end{align*} thus $f (x) \in O(g(x))$ as $x \to +\infty$, hence $f(n) \in O(g(n))$ as $\Bbb N^* \ni n \to \infty $.

For your question, it is simply the case $a = 10^{-6}$.

Answer 2

I don't have time to write a full answer, sry :c this should help: https://en.m.wikipedia.org/wiki/L%27Hôpital%27s_rule

Try to rewrite your fraction and then use l'Hôpital! :)

Answer 3

Use the standard limit $n^a/b^n\to 0$ as $n\to\infty$ for $b>1$. For your case $a=b=1.000001$ and thus the desired result is $0$.