 Define the sequence $f_1,f_2,\ldots :[0,1)\to \mathbb{R}$ of continuously differentiable functions by the following recurrence: $$ f_1=1; \qquad \quad f_{n+1}'=f_nf_{n+1} \quad\text{on $(0,1)$}, \quad \text{and}\quad f_{n+1}(0)=1. $$

Show that $\lim\limits_{n\to \infty}f_n(x)$ exists for every $x\in [0,1)$ and determine the limit function.
1. **Base Case and Induction Hypothesis:**
   - We start with the base case \( f_1(x) = 1 \). Clearly, \( f_1(x) \geq 1 \) for all \( x \in [0,1) \).
   - Assume \( f_n(x) \geq 1 \) for some \( n \geq 1 \). We need to show that \( f_{n+1}(x) \geq 1 \).

2. **Inductive Step:**
   - Given \( f_{n+1}' = f_n f_{n+1} \) and \( f_{n+1}(0) = 1 \), we can solve this differential equation.
   - Rearrange the differential equation:
     \[
     \frac{f_{n+1}'}{f_{n+1}} = f_n
     \]
   - Integrate both sides from 0 to \( x \):
     \[
     \int_0^x \frac{f_{n+1}'(t)}{f_{n+1}(t)} \, dt = \int_0^x f_n(t) \, dt
     \]
   - The left-hand side simplifies to:
     \[
     \ln |f_{n+1}(x)| - \ln |f_{n+1}(0)| = \int_0^x f_n(t) \, dt
     \]
   - Since \( f_{n+1}(0) = 1 \), we have:
     \[
     \ln f_{n+1}(x) = \int_0^x f_n(t) \, dt
     \]
   - Exponentiating both sides:
     \[
     f_{n+1}(x) = \exp \left( \int_0^x f_n(t) \, dt \right)
     \]
   - Since \( f_n(t) \geq 1 \) for all \( t \in [0,1) \) by the induction hypothesis, we have:
     \[
     \int_0^x f_n(t) \, dt \geq \int_0^x 1 \, dt = x
     \]
   - Therefore:
     \[
     f_{n+1}(x) \geq \exp(x) \geq 1
     \]
   - This completes the induction, showing \( f_n(x) \geq 1 \) for all \( n \geq 1 \).

3. **Monotonicity:**
   - We need to show that \( f_{n+1}(x) \geq f_n(x) \). We start with \( f_2(x) - f_1(x) = e^x - 1 \geq 0 \).
   - Assume \( f_{n+1}(x) \geq f_n(x) \). Then:
     \[
     \left( \frac{f_{n+1}}{f_n} \right)' = \frac{f_{n+1}' f_n - f_{n+1} f_n'}{f_n^2} = \frac{f_n f_{n+1} f_n - f_{n+1} f_n f_{n-1}}{f_n^2} = f_{n+1} \left( 1 - \frac{f_{n-1}}{f_n} \right)
     \]
   - Since \( f_n \geq f_{n-1} \), we have:
     \[
     \left( \frac{f_{n+1}}{f_n} \right)' \geq 0
     \]
   - This implies \( \frac{f_{n+1}}{f_n} \) is non-decreasing, so \( f_{n+1}(x) \geq f_n(x) \).

4. **Upper Bound:**
   - We show \( f_{n+1}(x) \leq \frac{1}{1-x} \). Consider:
     \[
     \left( \frac{1}{f_{n+1}} \right)' = -\frac{f_{n+1}'}{f_{n+1}^2} = -\frac{f_n f_{n+1}}{f_{n+1}^2} = -\frac{f_n}{f_{n+1}} \geq -1
     \]
   - Integrating from 0 to \( x \):
     \[
     \int_0^x \left( \frac{1}{f_{n+1}(t)} \right)' \, dt \geq \int_0^x -1 \, dt
     \]
   - This simplifies to:
     \[
     \frac{1}{f_{n+1}(x)} - \frac{1}{f_{n+1}(0)} \geq -x
     \]
   - Since \( f_{n+1}(0) = 1 \):
     \[
     \frac{1}{f_{n+1}(x)} \geq 1 - x
     \]
   - Therefore:
     \[
     f_{n+1}(x) \leq \frac{1}{1-x}
     \]

5. **Existence of the Limit:**
   - From steps 2 and 4, we have \( 1 \leq f_n(x) \leq \frac{1}{1-x} \) and \( f_n(x) \) is non-decreasing.
   - By the Monotone Convergence Theorem, \( \lim_{n \to \infty} f_n(x) = f(x) \) exists for every \( x \in [0,1) \).

6. **Determining the Limit Function:**
   - By the Dominated Convergence Theorem:
     \[
     f(x) = 1 + \lim_{n \to \infty} \int_0^x f_{n+1}'(t) \, dt = 1 + \lim_{n \to \infty} \int_0^x f_{n+1}(t) f_n(t) \, dt = 1 + \int_0^x f(t)^2 \, dt
     \]
   - Differentiating both sides:
     \[
     f'(x) = f(x)^2
     \]
   - Solving this differential equation with the initial condition \( f(0) = 1 \):
     \[
     \frac{df}{f^2} = dx \implies -\frac{1}{f} = x + C
     \]
   - Using \( f(0) = 1 \):
     \[
     -\frac{1}{1} = 0 + C \implies C = -1
     \]
   - Therefore:
     \[
     -\frac{1}{f} = x - 1 \implies f = \frac{1}{1-x}
     \]

The final answer is \( \boxed{\frac{1}{1-x}} \).