Does the graph the following curve have a tangent at the origin? So the first thing we're going to realize is we're going to use the formula limit is H goes to 0 of F of X + H -, F of X all over H And because we want it at the origin, X is going to equal 0 here. So we're going to have F We're going to have the limit as H goes to 0 of F of 0 + H -, F of 0 all over H Looking at our original conditions, AT0F of 0 is 0. So this end piece is just going to be 00 plus H is H. So this is really equivalent to limit as H goes to 0 of F of H / H So now we're talking about limit as H goes to 0, not at 0 as we get closer and closer and closer. So we're going to use that top equation and say H ^2 sine 1 / H all over H1 of the HS will cancel. So we get the limit as H goes to 0 of H times sine of 1 / H Well, if we stick an H going to 0 here 0 times anything is 0. Now sine. Because of what it does, we know that the sine values are always between one and -1, no matter what's inside. So if I have zero times something that's in between -1 and one, that's going to give us out zero. So yes, it exists because there is a tangent at that line. Or the limit as H goes to 0 of F of X + H -, F of X all over H exists.