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gowtham3214
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Hi. In lecture 3, in the derivation of the limit of (1cos Θ)/Θ as Θ>0 . I am able to understand tht the 'gap' 1cosΘ nears 0 as Θ tends to zero. But then, doesnt it mean we re actually ending up with 0/0 rather than 0. can anyone clarify on this ?
 3 years ago
 3 years ago
gowtham3214 Group Title
Hi. In lecture 3, in the derivation of the limit of (1cos Θ)/Θ as Θ>0 . I am able to understand tht the 'gap' 1cosΘ nears 0 as Θ tends to zero. But then, doesnt it mean we re actually ending up with 0/0 rather than 0. can anyone clarify on this ?
 3 years ago
 3 years ago

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alistair Group TitleBest ResponseYou've already chosen the best response.2
This is explained in Session 8 Course Notes Clip 3, Questions and Answers. Basically, as Θ tends to zero, the numerator (1cos Θ) shrinks faster than the denominator Θ, giving a limit of 0. (Think of Θ as the length of the arc here).
 3 years ago

gowtham3214 Group TitleBest ResponseYou've already chosen the best response.0
@alistair : thank you :) that helped :) clarified :)
 3 years ago

luckey Group TitleBest ResponseYou've already chosen the best response.0
do one thing multiply and divide the whole limit \[(1\cos \theta)\div \theta\] by an additional \[\theta\] then the read the limit as \[\lim_{\theta \rightarrow 0} ((1\cos \theta)\div \theta ^{2}) \theta\] solve the limit as \[\lim_{\theta \rightarrow 0} ((1\cos \theta)\div \theta ^{2}\] it'll come put to be 1/2 then the limit is left as \[\lim_{\theta \rightarrow 0} \theta\]*1/2 this is obviously equals to '0'
 2 years ago
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