Right triangle $ABC$ has right angle at $C$ and $\angle BAC=\theta$; the point $D$ is chosen on $AB$ so that $|AC|=|AD|=1$; the point $E$ is chosen on $BC$ so that $\angle CDE=\theta$.  The perpendicular to $BC$ at $E$ meets $AB$ at $F$.  Evaluate $\lim_{\theta\to 0}|EF|$.
1. **Identify the given information and setup the problem:**
   - Right triangle \(ABC\) with \(\angle C = 90^\circ\) and \(\angle BAC = \theta\).
   - Point \(D\) on \(AB\) such that \(|AC| = |AD| = 1\).
   - Point \(E\) on \(BC\) such that \(\angle CDE = \theta\).
   - Perpendicular from \(E\) to \(BC\) meets \(AB\) at \(F\).

2. **Determine the coordinates of points \(A\), \(B\), and \(C\):**
   - Let \(C\) be at the origin \((0,0)\).
   - Let \(A\) be at \((1,0)\) since \(|AC| = 1\).
   - Let \(B\) be at \((1, \tan \theta)\) since \(\angle BAC = \theta\) and \(|BC| = \tan \theta\).

3. **Find the coordinates of point \(D\):**
   - Since \(|AD| = 1\) and \(D\) lies on \(AB\), \(D\) must be at \((1,0)\) because \(|AD| = |AC| = 1\).

4. **Determine the coordinates of point \(E\):**
   - Point \(E\) lies on \(BC\) such that \(\angle CDE = \theta\).
   - Since \(\angle CDE = \theta\) and \(\angle DCE = 90^\circ - \theta\), \(E\) must be at \((1, \tan \theta)\).

5. **Find the coordinates of point \(F\):**
   - Point \(F\) is the intersection of the perpendicular from \(E\) to \(BC\) with \(AB\).
   - The slope of \(BC\) is \(\tan \theta\), so the slope of the perpendicular is \(-\cot \theta\).
   - The equation of the line through \(E\) with slope \(-\cot \theta\) is:
     \[
     y - \tan \theta = -\cot \theta (x - 1)
     \]
   - Simplifying, we get:
     \[
     y = -\cot \theta (x - 1) + \tan \theta
     \]
   - Since \(F\) lies on \(AB\), \(y = 0\):
     \[
     0 = -\cot \theta (x - 1) + \tan \theta
     \]
     \[
     \cot \theta (x - 1) = \tan \theta
     \]
     \[
     x - 1 = \tan^2 \theta
     \]
     \[
     x = 1 + \tan^2 \theta
     \]
   - Therefore, \(F\) is at \((1 + \tan^2 \theta, 0)\).

6. **Calculate the length \(EF\):**
   - The coordinates of \(E\) are \((1, \tan \theta)\) and \(F\) are \((1 + \tan^2 \theta, 0)\).
   - Using the distance formula:
     \[
     EF = \sqrt{(1 + \tan^2 \theta - 1)^2 + (\tan \theta - 0)^2}
     \]
     \[
     EF = \sqrt{(\tan^2 \theta)^2 + (\tan \theta)^2}
     \]
     \[
     EF = \sqrt{\tan^4 \theta + \tan^2 \theta}
     \]
     \[
     EF = \tan \theta \sqrt{\tan^2 \theta + 1}
     \]
     \[
     EF = \tan \theta \sec \theta
     \]
     \[
     EF = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta}
     \]
     \[
     EF = \frac{\sin \theta}{\cos^2 \theta}
     \]

7. **Evaluate the limit as \(\theta \to 0\):**
   \[
   \lim_{\theta \to 0} EF = \lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta}
   \]
   - Using L'Hospital's Rule:
     \[
     \lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta} = \lim_{\theta \to 0} \frac{\cos \theta}{-2 \cos \theta \sin \theta} = \lim_{\theta \to 0} \frac{1}{-2 \sin \theta} = \frac{1}{0} = \infty
     \]

However, the correct approach should be:
   \[
   \lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta} = \lim_{\theta \to 0} \frac{\sin \theta}{1} = 0
   \]

The final answer is \(\boxed{0}\)