Let $X_1,\dots, X_n$ be i.i.d. exponential random variables. For large $n$, what is the probability distribution for the following?
$$\frac{\max X_n}{\sum X_n}$$
I believe that the cdf of the maximum of $n$ i.i.d. exponential random variables is the $n$th power of the cdf of one when $n$ is fixed.
Write the statement as $\dfrac{\frac{\max X_n}{n}}{\frac{\sum X_n}{n}}$. Say the $X_n$'s are Exp($\lambda$).
The denominator $\frac{\sum X_n}{n}$ converges in probability (and distribution) to the mean of the $X_n$'s, $\frac{1}{\lambda}$ by the law of large numbers.
Now, you need to find the CDF of $\frac{\max X_n}{n}$ which is $P(\frac{\max X_n}{n}\leq c) = P(\max X_n \leq c n) = (1-e^{\lambda c n})^n$ for $c>0$, and $0$ otherwise (using the fact you noted). As $n \to \infty$, $P(\frac{\max X_n}{n}\leq c) \to 0$ if $c<0$ and $P(\frac{\max X_n}{n}\leq c) \to 1$ if $c>0$. Thus, $\frac{\max X_n}{n} \to 0$ in distribution (and thus in probability as well, since it converges to a constant).
Thus, by Slutksy's theorem, the limit is the ratio of the limits which is $0 / (1/\lambda) =0$ (i.e. a point mass at $0$).