You can use the substitution, let . If then , afterwards you should recognize a standard limit.
sin(h) / h as x-> 0 should be 0, but how do I formally show that? And why are we considering x->0 when we need x->infinity?
edit: after its sin(h)/h, can I use L'hopital rule and derive the top and bottom to get 1*cos(h) / 1, then then sub h as 1/x, then sub x as infinity, which results to 0, and cos of 0 is 1, so the limit is 1?
Or better yet, do a little research.
Think of the unit circle.
If the radius is one unit in length and angle made from the radius and the horizontal axis (measured in radians) is , then the green length is , the red length is , and the purple length is .
Clearly, the area of the smallest triangle is a little less than the area of the circular sector, which is a little less than the area of the larger triangle. So
It should be clear therefore, that .
(Actually, we have really only proven the right-hand limit, but the proof of the left-hand limit is almost identical.)