 One of the important limits to evaluate is sine x divided by x as x goes to zero. Let's find it numerically first, then try to verify our answer algebraically. There's just one problem we have to address. How are we measuring x? For example, suppose x is measured in degrees. Let's find the limit as x goes to zero of sine x divided by x numerically. And so we could try some values of x close to zero. And making sure our calculator is set to degree mode will calculate sine x divided by x. Now, all of our x values here were positive, so let's take a look at what happens if we use negative values. And there's certainly some trend here, and so this suggests that the limit as x approaches zero of sine of x divided by x is about 0.01745. On the other hand, suppose x is measured in radians. Let's go ahead and find that limit numerically. So again, we'll pick values of x close to zero, positive and negative, and evaluate sine x divided by x. And with our calculator set in radian mode, we find, and this suggests the limit as x approaches zero of sine x divided by x is a much nicer value, 1. And here's the important thing. The limit seems to depend on whether we're measuring in radians or in degrees. Now, since the radian limit is nicer, it's 1 instead of some decimal number. Let's always agree in calculus and beyond. Always measure angles in radians. So let's take a look at a limit. Suppose that theta is somewhere between zero and pi halves. So theta is a first quadrant angle, and let's go ahead and draw our unit circle with our angle theta. Again, if it's not written down, it didn't happen. Let's go ahead and label some points, our center O, and the point B with the terminal side of the angle intersects the circle, and this point X where the circle intersects the x-axis. And we get a couple of important figures. First of all, if we drop the perpendicular down to the x-axis and label, we get this triangle OAB. Next, we have the sector of the circle OXB. And finally, if our tangent extends to the terminal side of the angle, we get triangle OXY. Now we can see geometrically that triangle OAB is smaller than the sector OXB, and the sector OXB is smaller than the triangle OXY. And we can calculate these areas. Triangle OAB has area one-half base times height, and since this is the unit circle, our base is cosine theta, and our height is sine theta. And so the area will be one-half cosine theta sine theta. Now this sector, the area of a sector, is going to be proportional to the angle itself, and as long as we're measuring it radians, the area of that sector is going to be one-half theta. Remember this is a circle with a radius of one. And then finally this triangle OXY, the area is one-half base, that's one, times the height, and this height is tangent theta, and so the area, one-half, times one, times tangent theta. Now we can simplify this if we multiply everything by two that gets rid of that fraction, and if we then divide by sine of theta, we get, and so now I have theta divided by sine of theta squeezed in between two expressions, cosine theta, and one over cosine theta. Now if we want to stay in the first quadrant as theta goes to zero from above, our limits will be one for the outside function, which means that this middle function is squeezed in between the two of them, and so the limit as theta goes to zero from above, of theta divided by sine theta, must be equal to one. Now since nothing really changes if theta is negative, we also have the limit as theta goes to zero from below, theta divided by sine theta equal to one, and since the limit from the left and the limit from the right agree, it follows that the limit, no plus, no minus, is equal to one. Now one very important thing here, this limit is equal to one, but our inequality does not hold if we measure in degrees, and in particular this middle inequality, this area of the sector, is not going to be one half theta, but it's going to involve a constant multiplier. And if we wanted, we could find the limit if we measure angles in degrees, but why would we want to? Now we actually wanted to know the limit as x approaches zero of sine x divided by x, and the thing we know is that the limit as x approaches zero of sine x divided by x, well that's really the limit as x approaches zero of one divided by the reciprocal x divided by sine x. The limit of a quotient is the quotient of the limits, the numerator doesn't do anything interesting, and the denominator we just found that limit is equal to one, and so we arrive at an important conclusion. If theta is measured in radians, and it should be, then the limit as theta approaches zero of sine theta divided by theta is equal to one, and we can use this limit in a number of cases. So for example let's say we want to find the limit as x goes to zero of sine 3x divided by 5x. Now if we let theta equal 3x, then as x goes to zero, theta goes to zero as well, and so our limit can be rewritten, and while we could have done something with the 5x, let's leave it in place. Remember what we know about is the limit as x goes to zero of sine x divided by x. So we would like this to be sine theta divided by theta, and the thing to remember is you can get anything you want as long as you pay for it. So here we want a theta in the denominator, well that's a 3x, so let's just put a 3x in the denominator, and we'll pay for it by putting a 3x in the numerator. Now our 3x and our 5x multiply, so it doesn't matter which one goes first, so let's put the 3x in the first denominator and the 5x in the second. The limit of a product is the product of the limits. Theta is equal to 3x, and we can remove the common factor of x in our second limit, and now we have a familiar limit and a limit we can simplify, and our limit is three-fifths.