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ive been studying calculus nonstop for 6 hours today. only because my professor is a loser, so now i have to learn everything by myself. yay.
 

Determine the following limit.
lim x -> 0 [tan (3x)/ xcos(x)] = ?
does anyone know how to do this? i got 0 for answer, but i dunno if its right lol
 
It's actually pretty easy, this ones just a lil tricky. :(

this is what i have. i wish someone cud tell me if its right lol

lim x -> 0 [tan (3x)/ xcos(x)] = ?

lim x-> 0 sin3x/xcosx = 3[lim x->0 sin3x/3x] x [lim x ->0 1/cos3x] x [lim x -> 0 1/xcosx] = 3[1] x [1] x [1/1(0)] = 0. i dunno if that last zero is right.
:headache:
 
Sorry!

My food's teacher is like that. She's diabetic and all she does is talk about how she is one.

Then we're stuck learning on our own.. But it's only foods. So, I guess I can't really relate to your problem.
 
woohooo i went on yahoo!answers and got an answer.

i was vascillating between 0 and 3. lol


This may also be done using lim(x-->0) sin x / x = 1.
lim(x -> 0) tan (3x)/ [xcos(x)]
= lim(x -> 0) sin(3x)/ [***(3x) * xcos(x)]
= lim(x -> 0) [sin(3x)/ x] * 1/[***(3x) ***(x)]
= lim(x -> 0) 3 * [sin(3x)/(3x)] * 1/[***(3x) ***(x)]
= 3 * 1 * 1/1
= 3.

apparenttly the answer is 3 yay! im happy now :D

oh look at that. its almost like a work of art.

thank you random person for that haha
:)
 
i find it better doing it backwards.

insted of sin3x/xcosx, i got the answer right doing it xcosx/sin3x. and the answer i got i just reciprocated it. wooo,, i feel smart now.
:)
 
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