TITLE: Proving a limit to show an inequality, $\frac{\ln t}{t^{1/3}}$
QUESTION [5 upvotes]: This is a question on an assignment so please no full solutions, but if anyone could guide me through answering this question I'd be very greatful. Thanks
Show that $\displaystyle \frac{\ln t}{t^{1/3}} \rightarrow 0$ as $t \rightarrow \infty$. From the deﬁnition of limit (taking $\epsilon = 1$), deduce that for sufﬁciently large $t$, $(\ln t)
^3 \leq t$.

REPLY [2 votes]: When you have proved that $\lim_{t\to\infty}(\log t)/t^{1/3}=0$, by the definition of limit you can conclude that, for any $\varepsilon>0$, there exists $M$ so that, for any $t>M$, the inequality
$$
\left|\frac{\log t}{t^{1/3}}\right|\le \varepsilon
$$
holds.
It's not restrictive to assume that $M>1$, so from the inequality you get immediately
$$
\log t \le \varepsilon t^{1/3}
$$
or, equivalently,
$$
(\log t)^3 \le \varepsilon^3 t
$$
for all $t>M$.
What value of $\varepsilon$ do you need now?
So, what remains is to show that the limit is $0$, which shouldn't be too hard.

Note. Sorry, but I'm not able to write anything else than "$\log$" for the natural logarithm.