I'm stuck on the following problem and could use a hint:
Let $Z_1,\ldots,Z_n$ be IID random variables with density $f$. Suppose that $\mathbb{P}(Z_i > 0) = 1$ and that $\lambda = \lim_{x \to 0^{+} } f(x) > 0$. Let:
$$X_n = n\min\lbrace Z_1,\ldots,Z_n\rbrace$$
Show that $X_n \rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $\frac{1}{\lambda}$.
So far I've figured out that:
$$ \begin{align*} \mathbb{P}(X_n \leq x) &= \mathbb{P}(n\min\lbrace Z_1,\ldots,Z_n\rbrace \leq x) \\ &= \mathbb{P}(Z_1 \leq x/n \cup \ldots \cup Z_n \leq x/n) \\ &= 1 - \mathbb{P}(Z_1 > x/n \cap \ldots \cap Z_n > x/n) \\ &= 1 - \mathbb{P}(Z_1 > x/n) \cdot \ldots \cdot \mathbb{P}(Z_n > x/n) \\ &= 1 - \mathbb{P}(Z_1 > x/n)^n \\ &= 1 - (1-\mathbb{P}(Z_1 \leq x/n))^n \\ &= 1 - (1-F_{Z_1}(x/n))^n \end{align*} $$
But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,\ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,\ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.
If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n \to a \implies (1 + \frac{a_n}{n})^n \to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.
Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.
You need to prove this probability approximates $1-\exp -\lambda x$, i.e. that $(1-F_{Z_i}(x/n))^n\approx\exp -\lambda x$. But for $n\gg x$, $$F_{Z_i}(0)=0\implies F_{Z_i}(x/n)\sim \frac{x}{n} F_{Z_i}'(0)=\frac{\lambda x}{n},$$so the claim follows from $\exp -t=\lim_{n\to\infty}(1-\frac{t}{n})^n$.