Exercise about limit. Let be $g(t)=t\ln t$, $t>0$. How to show that $\displaystyle\lim_{t\rightarrow \infty} \frac{g^{-1}(t)}{t} = 0$ ?
I'm trying to do using the L'Hospital rule, but I can not justify that $\displaystyle\lim_{t\rightarrow \infty}g^{-1}(t)=\infty$. 
 A: The function $g$ is not globally invertible, but it is strictly increasing, hence invertible, on the interval $(e^{-1},\infty)$ and $g^{-1}$ must refer to this interval.
Since $\lim_{t\to\infty}g(t)=\infty$, also $\lim_{t\to\infty}g^{-1}(t)=\infty$, by the properties of invertible functions. Since $g$ is continuous (actually differentiable), also $g^{-1}$ is continuous (actually differentiable, because $g'$ is everywhere positive on $(e^{-1},\infty)$).
Now you can do the substitution $u=g^{-1}(t)$, so
$$
\lim_{t\to\infty}\frac{g^{-1}(t)}{t}=
\lim_{u\to\infty}\frac{u}{g(u)}=\lim_{u\to\infty}\frac{1}{\ln u}=0
$$
A: Note 1: If $f : I \to f(I)$ is strictly increasing on the interval $I \subset \mathbb{R}$, then so is its inverse $f^{-1} : f(I) \to I$. 
Proof:
Suppose $s < t$ with $f^{-1}(s) = x \in I$ and $f^{-1}(t) = y \in I$. Then $f(x) < f(y)$ by assumption, and since $f$ is strictly increasing on $I$, this implies $x < y$. Therefore $s < t \implies f(s) < f(t)$. 
Note 2:
If $f : I \to f(I)$ is invertible on $I$, then 
$$
\sup_{s \in f(I)}f^{-1}(s) = \sup(I) 
$$
Proof:
This is immediate since $f^{-1}(f(I)) = I$. 
From Here:
Put these pieces together to conclude that $g^{-1}(t) \to \infty$ as $t \to \infty$ in your case.
A: To show $g^{-1}(t)$ goes to infinity, you need to argue that if $y = t \ln t$ and $y$ gets large, then so does $t$.  Pick a large $N$.  If $y=N^2$, then $N \ln N < N^2 = y$, so $t> N$.  
Then for the limit, the inverse function theorem gives you $(g^{-1})^\prime(t) = 1/(1+\ln t).$  The derivative of the bottom is 1, so it's clear that the limit is $0$.
A: You get $g^{-1}(t) = \frac{t}{W(t)}$ where $W$ is the W-Lambert-function. Hence
$$\lim_{t \to \infty} \frac{g^{-1}(t)}{t} = \lim_{t \to \infty}\frac{1}{W(t)} = 0,$$
because $W(t) \to \infty$ for $x \to \infty$.
A: Without L'Hôpital:
Obviously $g^{-1}$ is always positive, hence the limit condition means that, for any $\varepsilon>0,$ $g^{-1}(t)<\varepsilon t$ if $t$ is large enough. So let us suppose there exists $\varepsilon>0$ s.t. $$g^{-1}(t)>\varepsilon t$$ for infinitely many $t$. Then we may consider $t>\frac1{\varepsilon e}$, and since $g$ is increasing on $(e^{-1},\infty)$ for infinitely many such $t$ we have $$g(g^{-1}(t))=t>\varepsilon t\log(\varepsilon t)=g(\varepsilon t),$$ i.e. $$\log t<\frac1\varepsilon-\log\varepsilon,$$ a contradiction.
