## Two Trigonometric Limits

We will show the two results

 1 h0 sin h h = 1 2 h0 cos h - 1 h = 0

Proof of (1)

First take a look at the following diagram, showing three areas arranged in order of magnitude.

The shaded area on the left (the smallest of the three) is a triangle with height of length sin h and base of length cos h. Therefore, its area is (cos h)(sin h)/2.

The pink shaded area (the next-smallest) is a circular segment compromising the fraction h/2 of the entire disc. Since the area of a disc of radius 1 is , the area in question is

 h2 . = h2

The shaded area on the right (the largest of the three) is a triangle with height of length tan h and base of length 1. Therefore, its area is (1)(tan h)/2 = (tan h)/2.

Putting these three areas therefore gives the inequality

 cos h sin h2 h2 tan h2 .

Writing tan h as the ratio (sin h)/(cos h) now gives

 cos h sin h2 h2 sin h2cos h .

Multiplying through by 2/(sin h) now gives

 cos h hsin h 1cos h .

Now take reciprocals and reverse inequalities to get

 1cos h sin h h cos h .

Finally, let h approach zero. As is does, the quantities on either end approach 1. Therefore, since the ratio (sin h)/h is sandwiched between two quantities approaching 1, it also approaces 1.

We are now done with the first limit we promised to compute.

Proof of (2)

For the second limit, we use a trigonometric identity and a little algebra:

h0
1 cos h

h
=  h0 1 cos hh . 1 + cos h1 + cos h
=  h0 1 cos2hh(1 + cos h)
=  h0 sin2hh(1 + cos h)
(using the identity sin2h + cos2h = 1)
=  h0 sinhh . sin h1 + cos h

The first term in this product is the limit we computed above, and has the value of 1. The second term approaches 0/(1+1) = 0. Therefore, the product approaches (1)(0) = 0, as required.