Here's the question you clicked on:
davetherave
how come is (sin (h)/h)=1...?
\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\] because the graph of y=x and sin x get closer and closer to each other as they approach zero. I assumed that's what you meant.
Use L'hopitals rule. If\[ \lim_{x \rightarrow a} \frac{ a(x) }{ b(x) }= \frac{ 0 }{ 0 }\] then \[\lim_{x \rightarrow a} \frac{ a(x) }{ b(x) }=\lim_{x \rightarrow a}\frac{ a(x) ' }{ b(x) ' }\]
To use L'Hopitals rule you need to know the derivative of sin(x), but to derivate sin(x) you need to know what is the limit of sin(x)/x when x approaches 0, so this is circular proof and thus not valid. Here is one proof that starts from the geometrical fact that if x is in radians and positive but less than pi/2 then sin(x) <= x <= tan(x). If we divide this inequity by sin(x) (>0) we get: 1 <= x/sin(x) <= 1/cos(x) If we take the inverse of this we get: 1>= sin(x)/x >= cos(x) If we now let x approach 0 we get: 1>= sin(0)/0 >= cos(0) So because cos(0) = 1 sin(0)/0 = 1.
Here's another interesting geometric proof using the squeeze theorem. http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx