Let $a_1\in (0,1)$ and $(a_n)_{n\ge 1}$ a sequence of real numbers defined by $a_{n+1}=a_n(1-a_n^2),\ (\forall)n\ge 1$. Evaluate $\lim_{n\to \infty} a_n\sqrt{n}$.
1. **Prove by induction that \(a_n \in (0,1)\):**
   - Base case: \(a_1 \in (0,1)\) is given.
   - Inductive step: Assume \(a_n \in (0,1)\). Then, \(a_{n+1} = a_n(1 - a_n^2)\).
     - Since \(a_n \in (0,1)\), \(a_n^2 \in (0,1)\).
     - Therefore, \(1 - a_n^2 \in (0,1)\).
     - Hence, \(a_{n+1} = a_n(1 - a_n^2) \in (0,1)\).
   - By induction, \(a_n \in (0,1)\) for all \(n \geq 1\).

2. **Show that \((a_n)_{n \geq 1}\) is decreasing:**
   - Consider \(a_{n+1} - a_n = a_n(1 - a_n^2) - a_n = -a_n^3\).
   - Since \(a_n \in (0,1)\), \(a_n^3 > 0\).
   - Therefore, \(a_{n+1} - a_n = -a_n^3 < 0\), implying that \((a_n)_{n \geq 1}\) is decreasing.

3. **Determine the limit \(a_n \to l\):**
   - Since \((a_n)_{n \geq 1}\) is decreasing and bounded below by 0, it converges to some limit \(l \in [0,1)\).
   - Taking the limit in the recurrence relation \(a_{n+1} = a_n(1 - a_n^2)\), we get:
     \[
     l = l(1 - l^2)
     \]
   - Solving \(l = l - l^3\), we get \(l^3 = 0\), hence \(l = 0\).
   - Therefore, \(a_n \to 0\).

4. **Apply the Stolz-Cesaro theorem:**
   - We need to evaluate \(\lim_{n \to \infty} a_n \sqrt{n}\).
   - Consider the sequence \(\frac{n}{\frac{1}{a_n^2}}\):
     \[
     \lim_{n \to \infty} \frac{n}{\frac{1}{a_n^2}} = \lim_{n \to \infty} n a_n^2
     \]
   - Using the Stolz-Cesaro theorem, we have:
     \[
     \lim_{n \to \infty} \frac{(n+1) - n}{\frac{1}{a_{n+1}^2} - \frac{1}{a_n^2}} = \lim_{n \to \infty} \frac{1}{\frac{1}{a_{n+1}^2} - \frac{1}{a_n^2}}
     \]
   - Since \(a_{n+1} = a_n(1 - a_n^2)\), we have:
     \[
     a_{n+1}^2 = a_n^2 (1 - a_n^2)^2
     \]
   - Therefore:
     \[
     \frac{1}{a_{n+1}^2} - \frac{1}{a_n^2} = \frac{1}{a_n^2 (1 - a_n^2)^2} - \frac{1}{a_n^2} = \frac{1 - (1 - a_n^2)^2}{a_n^2 (1 - a_n^2)^2} = \frac{2a_n^2 - a_n^4}{a_n^2 (1 - a_n^2)^2} = \frac{2 - a_n^2}{(1 - a_n^2)^2}
     \]
   - Thus:
     \[
     \lim_{n \to \infty} \frac{1}{\frac{2 - a_n^2}{(1 - a_n^2)^2}} = \lim_{n \to \infty} \frac{(1 - a_n^2)^2}{2 - a_n^2} = \frac{1}{2}
     \]
   - Hence, by the Stolz-Cesaro theorem:
     \[
     \lim_{n \to \infty} n a_n^2 = \frac{1}{2}
     \]
   - Therefore:
     \[
     \lim_{n \to \infty} a_n \sqrt{n} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}
     \]

The final answer is \(\boxed{\frac{\sqrt{2}}{2}}\).