Given distinct straight lines $ OA$ and $ OB$.  From a point in $ OA$ a perpendicular is drawn to $ OB$; from the foot of this perpendicular a line is drawn perpendicular to $ OA$.  From the foot of this second perpendicular a line is drawn perpendicular to $ OB$; and so on indefinitely.  The lengths of the first and second perpendiculars are $ a$ and $ b$, respectively.  Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds.  This limit is:

$ \textbf{(A)}\ \frac {b}{a \minus{} b} \qquad \textbf{(B)}\ \frac {a}{a \minus{} b} \qquad \textbf{(C)}\ \frac {ab}{a \minus{} b} \qquad \textbf{(D)}\ \frac {b^2}{a \minus{} b} \qquad \textbf{(E)}\ \frac {a^2}{a \minus{} b}$
1. Let's denote the angle between the lines \(OA\) and \(OB\) as \(\theta\). The first perpendicular from a point on \(OA\) to \(OB\) has length \(a\).
2. The second perpendicular from the foot of the first perpendicular to \(OA\) has length \(b\).
3. The third perpendicular from the foot of the second perpendicular to \(OB\) will be \(b \cos \theta\), where \(\cos \theta = \frac{b}{a}\).
4. The fourth perpendicular from the foot of the third perpendicular to \(OA\) will be \(b \cos^2 \theta\), and so on.

We observe that the lengths of the perpendiculars form a geometric sequence with the first term \(a\) and common ratio \(\cos \theta = \frac{b}{a}\).

5. The sum of an infinite geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)) is given by:
   \[
   S = \frac{a}{1 - r}
   \]
   Here, \(r = \frac{b}{a}\).

6. Substituting \(r\) into the formula, we get:
   \[
   S = \frac{a}{1 - \frac{b}{a}} = \frac{a}{\frac{a - b}{a}} = \frac{a \cdot a}{a - b} = \frac{a^2}{a - b}
   \]

Conclusion:
\[
\boxed{\frac{a^2}{a - b}}
\]