A soccer coach named $C$ does a header drill with two players $A$ and $B$, but they all forgot to put sunscreen on their foreheads.  They solve this issue by dunking the ball into a vat of sunscreen before starting the drill.  Coach $C$ heads the ball to $A$, who heads the ball back to $C$, who then heads the ball to $B$, who heads the ball back to $C$;
this pattern $CACBCACB\ldots\,$ repeats ad infinitum.  Each time a person heads the ball, $1/10$ of the sunscreen left on the ball ends up on the person's forehead.  In the limit, what fraction of the sunscreen originally on the ball will end up on the coach's forehead?
1. Let's denote the initial amount of sunscreen on the ball as \( S \).
2. Each time a person heads the ball, \( \frac{1}{10} \) of the remaining sunscreen on the ball is transferred to the person's forehead, and \( \frac{9}{10} \) of the sunscreen remains on the ball.
3. We need to find the fraction of the original sunscreen that ends up on Coach \( C \)'s forehead in the limit.

### Solution 1:
1. When Coach \( C \) heads the ball to \( A \), \( \frac{1}{10} \) of the sunscreen on the ball is transferred to \( C \)'s forehead. The ball retains \( \frac{9}{10} \) of the sunscreen.
2. The sequence of headers is \( CACBCACB\ldots \). Each time \( C \) heads the ball, \( \frac{1}{10} \) of the remaining sunscreen on the ball is transferred to \( C \)'s forehead.
3. The amount of sunscreen on \( C \)'s forehead after each cycle can be represented as an infinite geometric series:
   \[
   \frac{1}{10} \cdot S + \frac{1}{10} \cdot \left(\frac{9}{10}\right)^2 \cdot S + \frac{1}{10} \cdot \left(\frac{9}{10}\right)^4 \cdot S + \cdots
   \]
4. This series can be written as:
   \[
   \frac{1}{10} \cdot S \sum_{k=0}^{\infty} \left(\frac{9}{10}\right)^{2k}
   \]
5. The sum of the infinite geometric series is:
   \[
   \sum_{k=0}^{\infty} \left(\frac{9}{10}\right)^{2k} = \frac{1}{1 - \left(\frac{9}{10}\right)^2} = \frac{1}{1 - \frac{81}{100}} = \frac{1}{\frac{19}{100}} = \frac{100}{19}
   \]
6. Therefore, the fraction of the original sunscreen that ends up on Coach \( C \)'s forehead is:
   \[
   \frac{1}{10} \cdot \frac{100}{19} = \frac{10}{19}
   \]

### Solution 2:
1. Consider the repeating sequence \( \overline{CACB} \). We can treat the problem as if the repeating sequence is \( \overline{CA} \) because \( \overline{CACB} \) is equivalent to \( \overline{CACA} \).
2. After each cycle of \( CA \), the amount of sunscreen on the ball is reduced by a factor of \( \left(\frac{9}{10}\right)^2 \).
3. The ratio of the amount of sunscreen transferred to \( A \) and \( C \) is \( 9:10 \).
4. Therefore, the fraction of the original sunscreen that ends up on Coach \( C \)'s forehead is:
   \[
   \frac{10}{9 + 10} = \frac{10}{19}
   \]

The final answer is \(\boxed{\frac{10}{19}}\).