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anonymous
 5 years ago
Evalulate the following limit: limx>0 tanx/2x
anonymous
 5 years ago
Evalulate the following limit: limx>0 tanx/2x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have an indeterminate form of 0/0 and have to use l'Hopital's rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got = 1/2 by doing it normally

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really? cause \(lim_{x>0}tan x = 0\), and the \(\lim_{x>0} 2x \) is also 0. You cannot divide 0 by 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh no not that way I meant the way in my notes: =limx>0 1/2 (sinx/cosx)/x = 1/2 limx>0 sinx/xcosx = 1/2 (limx>0 sinx/x)(limx>0 1/cosx) = 1/2 (1)(1) =1/2 Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you can get there by L'hopital, but this is one of those exercise that they always throw in Cal I. Recognize \[((\tan x)/x)=1\] So pull the 1/2 out front of the lim sign. So as the tan thingy goes to 1, what's left is 1/2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It makes sense as long as you already know that the limit as x goes to 0 of sinx/x is 1
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