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NoelGreco
 2 years ago
Best ResponseYou've already chosen the best response.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.

snuderl
 2 years ago
Best ResponseYou've already chosen the best response.0Use L'hopitals rule. If\[ \lim_{x \rightarrow a} \frac{ a(x) }{ b(x) }= \frac{ 0 }{ 0 }\] then \[\lim_{x \rightarrow a} \frac{ a(x) }{ b(x) }=\lim_{x \rightarrow a}\frac{ a(x) ' }{ b(x) ' }\]

Topi
 one year ago
Best ResponseYou've already chosen the best response.0To 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.

dinnertable
 one year ago
Best ResponseYou've already chosen the best response.0Here's another interesting geometric proof using the squeeze theorem. http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx
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