 In this video, we will provide the solution to question number two from practice exam number three for math 1210. We're asked to compute the limit as theta approaches zero of sine theta over five theta squared minus four theta. So my first inclination is that since everything looks continuous, what just happens if I plug in theta equals zero, just kind of like test the water a little bit, you're going to get sine of zero over five times zero squared minus four times zero. For which sine of zero is zero in the denominator, you're going to get zero minus zero, so you end up with zero over zero. This is an indeterminate form. This does not tell you that the limit doesn't exist. I mean, that could be the case, but we don't have enough information. We have to try something else. So what we know about sine here is the very following observation. If we take the limit as theta approaches zero of sine theta over theta, this we learn from the squeeze theorem is equal to one. So can I somehow make that pop up in this limit somehow? Well, notice that theta is actually a factor of the denominator. If we factor it out, we end up with the limit as theta approaches zero of the denominator. Let's do that one first. We get theta times five theta minus four. This sits above sine of theta. So notice now how we have the sine theta over theta that has now showed up in our limit process. That means that's going to turn out to be a one. So we get the limit of sine theta over theta. Take the limit as theta goes to zero. And we're multiplying that by the limit as theta goes to zero of one over five theta minus four. For which on the first one, you see that this limit is going to go off towards one. And in the second limit, if we plug in theta equals zero, we don't get the indeterminate form we had anymore. We end up with one times one over zero minus four. This becomes negative one fourth. And so we see that the correct answer is E.