1/10 is a periodic number in binary, just like 1/3 is a periodic number in decimal. As such, it can't be accurately stored as a floating-point number.

First, reduce the accumulation of error through the use of integers.

Secondly, use rounding or a tolerance when appropriate.

for my $i (0..89) {
   my $v = $i / 10;
   printf "%.1f\n", $v;
}
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(Because we avoided the accumulation of error, the rounding isn't actually needed here.)