## 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).

• What is the area of triangle ABC? This is (1/2)base·height. The base is AB, which is a radius of the circle, so the length of AB is 1. The height is CD, the opposite side from the angle with measurement h and hypotenuse AC (another unit length, since another radius of the circle). Therefore sin(h) is the quotient of the length of CD divided by 1. So CD has length sin(h), and the area of triangle ABC is (1/2)sin(h)·1.
• What is the area of triangle ABE? This is (1/2)base·height. The base is AB, which is a radius of the circle, so the length of AB is 1.The height is EB, but EB/AB is tan(h), and the length of AB is 1, so EB has length tan(h). The area of triangle ABE is (1/2)tan(h)·1.
• What is the area of sector ABC? Here we need to think a bit. Some examples of sectors of circles and their areas may be useful:
The area of a whole circle of radius 7 is 72\Pi.
The area of a half circle of radius 5 is 52P\i/2.
The area of a quarter circle of radius 3 is 32Pi/4.
The point of this was to convince you that the area of a sector is directly proportional to the product of the square of the radius and the angle measurement (in radians). The constant of proportionality is (1/2): that is, the area of a circular sector of radius R and central angle Theta is (1/2)R2Theta.
Therefore the area of circular sector ABC is (1/2)12h=(1/2)h. (There's no Pi in this answer, because the Pi is already incorporated into the radian measure of the central angle!)
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
(1/2)sin(h)(1/2)h(1/2)tan(h)

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.