Verification that the limit of sin(h)/h as h-->0 is 1

Here is a discussion supporting the assertion that the stated limit is 1. The discussion needs pictures. To me this is good since I am a picture person.

Look at the accompanying picture to the right. This picture shows a very small angle h, inside the unit circle (all of the radii are equal to 1).

Now we see which of the areas is largest and which is smallest and which is middlest. (The sequence of letters "middlest" does not seem to be an English word. I am sorry.)

Smallest areaMiddlest areaLargest area
Triangle ABCSector ABDTriangle ABE

The first two entries in the last row imply that (1/2)sin(h)&le(1/2)h. Let's multiply by 2 and divide by h. Then [sin(h)/h]≤1. The last two entries in the last row give us (1/2)h≤(1/2)tan(h). Well, tan(h)=sin(h)/cos(h), so we have (1/2)h≤(1/2)sin(h)/cos(h). Now multiply by 2 (gets rid of the 1/2's) and divide by h and multiply by cos(h). The result is cos(h)≤sin(h)/h.

Let's put the inequalities we have together:   cos(h)<=[sin(h)/h]<=1. We are interested in what happens as h-->0. Well, here is a valid use of the squeeze theorem, since both cos(h) and 1 approach 1 as h-->0. So we can finally conclude that limh-->0[sin(h)/h]=1.

For example, on my "calculator", I just asked for sin(.0123) and got 0.0002146755. WHAT!!!??? Isn't this wrong? Isn't this way off? Well, no. I actually asked the calculator the wrong question. The calculator was set for degrees, not for radians. If you insist that your trig functions are functions of degrees, then the derivatives will be all fouled up. In fact, the true value of sine of .0123 is actually 0.01229969, which is pretty darn close. So sine of h radians is quite close to h when h is small.

Maintained by and last modified 7/2/2006.