Proving a limit to show an inequality, $\frac{\ln t}{t^{1/3}}$ 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 $t$, $(\ln t)
^3 \leq t$.
 A: ADD If you happen not to know this:


*

*By an injection from a set $A$ to a set $B$ is meant a function $f:A\to B$ which is one-one, or injective.

*By a monotone function we mean a function which "preserves" or "inverses" order always. That is, either $a\leq b$ implies $f(a)\leq f(b)$ (always) or $a\leq b$ implies $f(a)\geq f(b)$ always.

Note that the map $$x\mapsto x^3$$ is an injection of positive reals to positive reals, that preserves monotonicity. If we look at your limit $$\lim_{t\to \infty}\frac{\log t}{t^{1/3}}=\ell$$
we can argue this is the same as letting $t^3\to \infty$, so that  $$\lim_{t\to \infty}\frac{\log t^3}{t}=\ell$$
or
we can argue this is the same as letting $t^3\to \infty$, so that  $$\lim_{t\to \infty}3\frac{\log t}{t}=\ell$$
Now, we can inject $\Bbb R^+$ onto $\Bbb R_{\geq 1}$ by $t\mapsto e^t$, and preserving monotonicity (i.e. $t\to\infty \iff e^t \to \infty$) so that
$$\lim_{t\to \infty}3\frac{\log e^t}{e^t}=\ell$$
which gives
$$\lim_{t\to \infty}\frac{3t}{e^t}=\ell$$
Can you show that given any $\epsilon >0$ there exists $M>0$ such that for $t\geq M$, $te^{-t}<\epsilon$?

From the definition of the limit, $t^{-1/3}\log t \to 0$ as $t\to \infty $ means that for every $\epsilon >0$ there exists $M>0$ such that for $t\geq M$; we have $t^{-1/3}\log t <\epsilon$. In particular, if $\epsilon =1$ is given, there exists an $M'$ such that for $t\geq M$ - that is, over $[M,\infty)$ - we have $$t^{-1/3}\log t <1$$ 
 which is the same as...?
A: 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.
