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Does anyone know what the answer is to this question?
The limit of X --> 0 degrees of (Sin x)/X
any help would be appreciated.
are you allowed to use l'Hopital's rule?
A logarithm is just a misspelled algorithm.
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Yes, if makau's question is answered yes, the answer is 1. Because d/dx of Sinx is Cosx and d/dx of x is 1.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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thats the answer? i dont know, the question is in degrees not radians
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zero degrees or zero radians, both are equal.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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no, the answer is not 1...look for an article by John Loase.
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this limit is one of famous limits
you can solve this limit by l'Hopital's rule(if you don't know this rule tell me to explain it).
The limit of X --> 0 degrees of (Sin x)/X = 1
The limit of X --> 0 degrees of (tan x)/X = 1
The limit of X --> 0 degrees of (1-cosx)/X = 0
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it's not 1....i have to find the article, but it's not 1
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Dude, it is 1. Try sin(.1)/.1 and sin(0.001)/0.001 and ull see that it is approaching 1. Try the negative values and ull see ull get the same approach. Clearly it is 1.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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no..it's not, i will prove it as soon as i get the article.
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