Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four.