How your prove "

" depends, of course, upon exactly how you have

**defined** sin(x).

We can, for example,

**define** sin(x) to be the power series

. Then it is easy to see that

which clearly has limit 1 as x goes to 0.

Or we can define sin(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 0, y'(0)= 1" while defining cos(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 1, y'(0)= 0".

Using those definitions, if we let y= (sin(x))', the derivative of sin(x), then y'= ((sin(x))')'= (sin(x))''= -(sin(x)) so that y''= -(sin(x))'= -y. That is, this new function also satisfies the equation y''= -y. Further y(0)= (sin(x))' at 0 which is 1 and y'(x)= (sin(x))''= - sin(x) is 0 at x= 0. By "uniqueness" of the solutions of these initial value problems, then, y(x)= cos(x). That is (sin(x))'= cos(x) and, since we have got that derivative without using the limit, we can use L'Hopital's rule:

.