anonymous
  • anonymous
Evalulate the following limit: limx->0 tanx/2x
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
You have an indeterminate form of 0/0 and have to use l'Hopital's rule.
anonymous
  • anonymous
I got = 1/2 by doing it normally
anonymous
  • anonymous
Really? cause \(lim_{x->0}tan x = 0\), and the \(\lim_{x->0} 2x \) is also 0. You cannot divide 0 by 0.

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anonymous
  • anonymous
Oh 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
  • anonymous
I 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
  • anonymous
It 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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