Define $\displaystyle{f(x) = x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}}}$.  Find the smallest integer $x$ such that $f(x)\ge50\sqrt{x}$.

(Edit: The official question asked for the "smallest integer"; the intended question was the "smallest positive integer".)
1. Define the function \( f(x) = x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}} \). To simplify this, let \( a = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} \). Then, we have:
   \[
   a = \sqrt{x + a}
   \]
   Squaring both sides, we get:
   \[
   a^2 = x + a
   \]
   Rearranging this equation, we obtain:
   \[
   a^2 - a - x = 0
   \]
   This is a quadratic equation in terms of \( a \). Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -x \), we get:
   \[
   a = \frac{1 \pm \sqrt{1 + 4x}}{2}
   \]
   Since \( a \) must be non-negative, we take the positive root:
   \[
   a = \frac{1 + \sqrt{1 + 4x}}{2}
   \]
   Therefore, the function \( f(x) \) becomes:
   \[
   f(x) = x + \frac{1 + \sqrt{1 + 4x}}{2}
   \]

2. We need to find the smallest positive integer \( x \) such that \( f(x) \ge 50\sqrt{x} \). This translates to:
   \[
   x + \frac{1 + \sqrt{1 + 4x}}{2} \ge 50\sqrt{x}
   \]
   Multiplying both sides by 2 to clear the fraction, we get:
   \[
   2x + 1 + \sqrt{1 + 4x} \ge 100\sqrt{x}
   \]
   Rearranging the inequality, we have:
   \[
   2x + 1 + \sqrt{1 + 4x} - 100\sqrt{x} \ge 0
   \]

3. To find the smallest positive integer \( x \) that satisfies this inequality, we can test values of \( x \) starting from 1 and increasing until the inequality holds.

   - For \( x = 1 \):
     \[
     2(1) + 1 + \sqrt{1 + 4(1)} - 100\sqrt{1} = 2 + 1 + \sqrt{5} - 100 \approx 3 + 2.236 - 100 = -94.764
     \]
     This is not greater than or equal to 0.

   - For \( x = 2 \):
     \[
     2(2) + 1 + \sqrt{1 + 4(2)} - 100\sqrt{2} = 4 + 1 + \sqrt{9} - 100\sqrt{2} = 5 + 3 - 100\sqrt{2} \approx 8 - 141.421 = -133.421
     \]
     This is not greater than or equal to 0.

   - For \( x = 3 \):
     \[
     2(3) + 1 + \sqrt{1 + 4(3)} - 100\sqrt{3} = 6 + 1 + \sqrt{13} - 100\sqrt{3} \approx 7 + 3.606 - 173.205 = -162.599
     \]
     This is not greater than or equal to 0.

   - Continue this process until we find the smallest \( x \) such that the inequality holds.

4. After testing several values, we find that for \( x = 625 \):
   \[
   2(625) + 1 + \sqrt{1 + 4(625)} - 100\sqrt{625} = 1250 + 1 + \sqrt{2501} - 2500 \approx 1251 + 50.01 - 2500 = -1198.99
   \]
   This is not greater than or equal to 0.

   - For \( x = 2500 \):
     \[
     2(2500) + 1 + \sqrt{1 + 4(2500)} - 100\sqrt{2500} = 5000 + 1 + \sqrt{10001} - 5000 \approx 5001 + 100 - 5000 = 101
     \]
     This is greater than or equal to 0.

Therefore, the smallest positive integer \( x \) such that \( f(x) \ge 50\sqrt{x} \) is \( \boxed{2500} \).