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**HallsofIvy** How your prove "$\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}$" depends, of course, upon exactly how you have **defined** sin(x).

We can, for example, **define** sin(x) to be the power series $\displaystyle x- x^3/3!x+ \cdot\cdot\cdot= \sum_{n=0}^\infnty \frac{x^{2n+1}}{(2n+1)!}$. Then it is easy to see that $\displaystyle \frac{sin(x)}{x}= \frac{x- x^3/6+\cdot\cdot\cdot}{x}= 1- x^2/6+ \cdot\cdot\cdot$ 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: $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}= \lim_{x\to 0}\frac{cos(x)}{1}= 1$.