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anonymous
 4 years ago
In sinθ/θ, I see that as θ approaches zero both approach zero however what I do not see proven is that sinθ approaches zero faster than θ.
anonymous
 4 years ago
In sinθ/θ, I see that as θ approaches zero both approach zero however what I do not see proven is that sinθ approaches zero faster than θ.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We often prove the limit as theta is approaching zero of the sine of theta divided by theta is equal to one using the Squeeze Theorem or sometimes called the Sandwich Theorem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327342370121:dw dw:1327342739125:dw dw:1327342828759:dw dw:1327343029234:dw dw:1327343164640:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I may not have understood your question....I assumed you were asking why the limit was equal to one. Your statement that sine of theta is approaching zero "faster" than theta is approaching zero....comes from what source?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327344166449:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You have shown me what puzzled me. I was looking at the blackboard diagrams showing in Lect. 3. Yes that is the diagram, and though intuitively sin theta is smaller than theta, I guess I was looking for something nonvisual like your squeeze theorem. I do not trust my intuition yet.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah, okay...I hope it helped...have a good one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The squeeze theorem and your diagram helped me unfreeze my mind; helping me see that I had been ignoring that the arc and the angle were equal.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help.
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