Evaluating $\lim_{n\to \infty} \frac{1}{\theta} (1-\frac{z/\theta}{n})^{n-1}$

Solution 1:

I would go for CDF's at once (BTW it stands for "cumulative distribution function")

First of all let us do it for iid $Y_1,Y_2,\dots$ that have standard uniform distribution (i.e. $\theta=1$ in this context). This to avoid the in my view annoying parameters.

Then for $x>0$ and $n$ large enough: $$F_n(x)=P\left(n\left(1-Y_{(n)}\right)\leq x\right)=1-P\left(Y_{(n)}<1-\frac{x}{n}\right)=1-\left(1-\frac{x}n\right)^n$$

This reveals that $\lim_{n\to\infty}F_n(x)=F(x)$ for every $x\in\mathbb R$ where $F$ denotes the CDF of standard exponential distribution.

This justifies the statement $$n\left(1-Y_{(n)}\right)\stackrel{d}{\to}Z$$where $Z$ has standard exponential distribution.

Now take $X_i:=\theta Y_i$ for $i=1,2,\dots$ and it follows immediately that $$n\left(\theta-X_{(n)}\right)\stackrel{d}{\to}\theta Z$$

Here $\theta Z$ has exponential distribution with mean $\theta$.

Solution 2:

$$\lim_{n\to \infty} \frac{1}{\theta} \left(1-\frac{z/\theta}{n}\right)^{n-1}=\frac{1}{\theta}\lim_{n\to \infty}[\frac{\left(1-\frac{z/\theta}{n}\right)^{n}}{\left(1-\frac{z/\theta}{n}\right)}]=\frac{1}{\theta}\cdot\frac{\lim_{n\to \infty}\left(1-\frac{z/\theta}{n}\right)^{n}}{\lim_{n\to \infty}\left(1-\frac{z/\theta}{n}\right)}=\frac{1}{\theta}\cdot\frac{e^{-z/\theta}}{1}=\frac{1}{\theta}e^{-z/\theta}$$

$$Q.E.D.$$