For this problem, we're going to stick in seven Pi eighteenths, so the limit as X approaches 7 Pi eighteenths of cosine 9X minus cosine of 9X. So when we stick in seven Pi eighteenths, we're going to get cosine of 9 * 7 Pi eighteenths minus cosine of 9 * 7 Pi eighteenths. Well, 9 * 7 Pi eighteenths is going to give me 7 Pi halves, because remember that nine is understood over 19 and 18 reduced to 1/2 minus cosine of. Once again, this one's going to reduce to seven Pi halves. What is cosine of seven Pi halves? Cosine of any K π half is 0, so we're going to have cosine 7 Pi halves -0 seven Pi halves -0 is going to be cosine of seven Pi halves again, and cosine of any K Pi halves is going to be 0. Actually I should say any K Pi halves where K is an odd integer because 2π halves would be a π and cosine of π is not zero. So Pi halves three Pi halves 5 Pi halves -1 half Pi -3 halves Pi, etcetera. So that's our final answer. There was a Part 2 of this problem and the Part 2 wants to know is this continuous? And so we want to know if I put in seven Pi eighteens without the limit portion, is that location defined? And our answer is yes, because we just did a straight substitution. We didn't have to use a conjugate or a theorem or anything else to reduce it down. So yes, it is continuous at this location.