Last night my instructor was demonstrating the squeeze theorem and he used a problem that he was unable to figure out. I’m assuming he’ll have it worked out by tomorrow night’s class. Basically, it was:
Calculate the limit of sin theta/theta as theta approaches zero.
Using a bit of geometry and arc lengths he showed that theta will always be a bit larger than sin theta, (sin theta < theta) and dividing by theta gave us an upper bound of 1. However he couldn’t recall the correct formula for the lower bound, though he assures us the answer for the limit is 1. I can’t think of what the lower bound could be and it’s really nagging me. I suppose it could be 0 but then the squeeze theorem wouldn’t work, since 0/theta will be 0 (duh). Is this some huge, involved proof?
Now note that tan(th) > th. To prove this, consider the area of sector of the unit circle subtended by angle th; this must be .5th. It’s easy to show that the right triangle formed by the unit base along the x-axis (line from (0,0) to (1,0)) with right angle at (1,0) and angle th at the origin has an area of .5tan(th) (wish I had a drawing…). In any event, the area of this right triangle contains the area of the sector plus a bit more, so tan(th) > th.
So sin(th) < th < tan(th). Eventually you get cos(th) < sin(th)/th < 1, and since lim cos(th = 1 as th goes to zero, the squeeze theorem gives you the limit.
That’s just wrong. If you work in radians (which are, essentially, God’s units) then you get 1. However, if theta is measured in degrees then you’ll get something like pi/180. In neither case will you get zero.
Strictly speaking, you still have units in the denominator, so you’d get pi/180 degrees. But of course, 180 degrees is equal to pi, so you still get 1.
Note that (in my old Calc book) the limit in the OP was used to prove the derivative of sin x. Hence any method that involves taking a derivative of sin x leads to a circular argument (at least in that context). E.g., using L’Hopital’s Rule or the Taylor series expansion of sin x.