GET your dubs together
Mathematica, 29 bytes

How about an arithmetic solution?

IntegerExponent[9#+#~Mod~10]&amp;


I'm very pleased to see that this beats the straight-forward Mathematica approach.

Explanation

The code itself computes 9*n + n%10 and then finds the largest power of 10 that divides the input, or in other words, counts the trailing zeros. We need to show if n ends in k repeated digits, that 9*n + n%10 has k trailing zeros.

Rep-digits are most easily expressed mathematically by dividing a number like 99999 (which is 105-1) by 9 and then multiplying by the repeated digit. So we can write n = m*10k + d*(10k-1)/9, where m ≢ d (mod 10), to ensure that n doesn't end in more than k repeated digits. Note that d = n%10.

Let's plug that into our formula 9*n + n%10. We get 9*m*10k + d*(10k-1) + d. The d at the end is cancelled, so we're left with: 9*m*10k + d*10k = (9*m + d)*10k. But 9 ≡ -1 (mod 10), so 9*m + d ≡ d - m (mod 10). But we've asserted that m ≢ d (mod 10) and hence d - m ≢ 0 (mod 10). 

In other words, we've shown that 9*m + d is not divisible by 10 and therefore, the largest power of 10 that divides 9*n + n%10 = (9*m + d)*10k is k, the number of trailing repeated digits.

As a bonus, this solution prints the correct result, ∞, for input 0.