Question

Evalulate the following limit: limx->0 tanx/2x

Answer 1

You have an indeterminate form of 0/0 and have to use l'Hopital's rule.

Answer 2

I got = 1/2 by doing it normally

Answer 3

Really? cause \(lim_{x->0}tan x = 0\), and the \(\lim_{x->0} 2x \) is also 0.
You cannot divide 0 by 0.

Answer 4

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?

Answer 5

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.

Answer 6

It makes sense as long as you already know that the limit as x goes to 0 of sinx/x is 1