$k$ Numbers are chosen from the set $1-n$ with replacement. Express: All $k$ are distinct for $n → ∞$ and $\frac{\sqrt{k}}{n}→ θ$ Numbers $ X_1, X_2,  ... , X_k $ are chosen randomly from the set {$ 1, 2, . . . , n$}
with replacement.
(a) Find the probability that all chosen numbers are distinct.


*

*Solved: $$ \frac{n\cdot(n-1)\cdot(n-2)\cdot...\cdot(n-k+1)}{n^k} $$


(b) Find the limit of the probability calculated in part (a) if $n →
∞$ and $\frac{\sqrt{k}}{n}→ θ$ for some $ 0 < θ < ∞$.
(Hint: For sufficiently small $x \ge 0$ we have: $ e^{-x-x^2} \le 1-x \le e^{-x}  $)
I started section (b) by presenting the probaility I've found at the previous section as:
$ P(A)=\prod_{l=1}^{k-1}{(1-\frac{l}{n})} $
and I didn't know how to proceed.
I could use some directions, thanks in advance.

As Sergei pointed out in the comments, $\sqrt{k}/n \to \theta$ does not make sense.
Perhaps you meant $k/\sqrt{n} \to \theta$.
Assume in the limit, the inequality in the hint holds for $x=l/n$ for each $l$ in your product. [This would hold if $k/\sqrt{n} \to \theta$, since then we would have $0<l/n < k/n \to 0$.]
Then the probability is bounded as
$$\exp\left(-\sum_{l=1}^{k-1} \left(\frac{l}{n}+\frac{l^2}{n^2}\right)\right)
\le \prod_{l=1}^{k-1} \left(1-\frac{l}{n}\right) \le \exp \left(-\sum_{l=1}^{k-1} \frac{l}{n}\right)$$
$$e^{-\frac{k(k-1)}{2n} - \frac{(k-1)k(2k-1)}{6n^2}}\le P(A) \le e^{-\frac{k(k-1)}{2n}}$$
Taking limits gives $P(A) \to e^{-\theta/2}$.
