lim x approach 0 (cot 2x)/(csc x)
2007-04-11 17:41:03 · 4 answers · asked by diYana = 1 in Science & Mathematics ➔ Mathematics
(cot 2x)/(csc x) can be rewritten as (sin x)/(tan 2x) sin 0=0 and tan 0=0 so u use L'Hopitals rule and you get (cos x)/(2 sec^2 2x)) and when you plug in 0 for x you get 1/2..... I don't know what the sandwich theorem is tho, sry...
2007-04-11 17:49:32 · answer #1 · answered by chosen_one_of_atlantis 1 · 0⤊ 0⤋
cot 2x / csc x = cos 2x . sin x / sin 2x
= cos 2x sin x / (2 sin x cos x)
= cos 2x / (2 cos x).
So lim (x->0) (cot 2x / csc x)
= lim (x->0) cos 2x / (2 cos x)
= 1/2.
There's no need to use the sandwich (squeeze) theorem at all.
2007-04-12 00:52:17 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
Sandwich? I think you mean squeeze theorem, but I could be wrong. I've never heard of a sandwich theorem though, sorry!
(I'd go ahead and try to do it using squeeze theorem for you, but I also hate trig and am terrible at it, so you'd pretty much be dumber after having looked at my work)
2007-04-12 00:50:08 · answer #3 · answered by macho_bob 3 · 0⤊ 0⤋
You'll always be hungrary
2007-04-12 00:57:49 · answer #4 · answered by calpal2001 4 · 0⤊ 0⤋