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NoelGrecoBest 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.
 one year ago

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

TopiBest ResponseYou've already chosen the best response.0
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.
 one year ago

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