Let $x_1=2021$, $x_n^2-2(x_n+1)x_{n+1}+2021=0$ ($n\geq1$). Prove that the sequence ${x_n}$ converges. Find the limit $\lim_{n \to \infty} x_n$.
1. **Define the sequence and the function:**
   Given the sequence \( \{x_n\} \) with \( x_1 = 2021 \) and the recurrence relation:
   \[
   x_n^2 - 2(x_n + 1)x_{n+1} + 2021 = 0 \quad \text{for} \quad n \geq 1
   \]
   We can rewrite the recurrence relation in terms of a function \( f \):
   \[
   x_{n+1} = f(x_n) = \frac{x_n^2 + 2021}{2(x_n + 1)}
   \]

2. **Identify the fixed point:**
   Let \( a = 2021 \) and \( \alpha = \sqrt{a + 1} - 1 \). We need to show that \( \alpha \) is a fixed point of \( f \):
   \[
   f(\alpha) = \frac{\alpha^2 + 2021}{2(\alpha + 1)}
   \]
   Substituting \( \alpha = \sqrt{2022} - 1 \):
   \[
   f(\sqrt{2022} - 1) = \frac{(\sqrt{2022} - 1)^2 + 2021}{2(\sqrt{2022} - 1 + 1)} = \frac{2022 - 2\sqrt{2022} + 1 + 2021}{2\sqrt{2022}} = \frac{2022}{2\sqrt{2022}} = \sqrt{2022} - 1 = \alpha
   \]
   Thus, \( \alpha \) is indeed a fixed point of \( f \).

3. **Monotonicity and boundedness:**
   To show that \( f \) is strictly increasing on \( [\alpha, \infty) \), consider the function:
   \[
   g(x) = \frac{x^2 + 2021}{x + 1} = x - 1 + \frac{2022}{x + 1}
   \]
   Compute the derivative:
   \[
   g'(x) = 1 - \frac{2022}{(x + 1)^2}
   \]
   For \( x > \alpha \), we have \( (x + 1)^2 > 2022 \), hence:
   \[
   g'(x) = 1 - \frac{2022}{(x + 1)^2} > 0
   \]
   Therefore, \( g(x) \) and consequently \( f(x) \) is strictly increasing on \( [\alpha, \infty) \).

4. **Behavior of the sequence:**
   For \( x > \alpha \), we have:
   \[
   \alpha < f(x) < x
   \]
   This implies that the sequence \( \{x_n\} \) is decreasing and bounded below by \( \alpha \). Hence, \( x_n \) converges to some limit \( t \geq \alpha \).

5. **Finding the limit:**
   Since \( x_n \to t \) and \( f(t) = t \), we have:
   \[
   t = \alpha
   \]
   Therefore, \( x_n \) converges to \( \alpha \).

The final answer is \( \boxed{ \lim_{n \to \infty} x_n = \sqrt{2022} - 1 } \).