<p>Find the limit of x as it approaches 0. [3x+(1-(cos x)^2)]/[sin x]</p>
<p>I have no idea what I am doing on this problem...</p>
<p>I thought it was a straightforward plug in sinx^2 for (1-cosx^2) but of course not...lol</p>
<p>Any help would be appreciated.</p>
<p>I think you got the first part right. I would make it (pretend there's a as the limit approaches 0 infront of all my steps :)
(3x+sin^2x)/sinx
I would then seperate it into 2 different equations
lim 3x/sinx + lim sin^2x/sinx
3lim x/sinx (which = 1) +lim sinx
3 + 0
3
is that the correct answer?</p>
<p>o and i got that x/sinx = 1 because of it's reciprocal. since sinx/x = 1 the reciprocal would also equal 1.</p>
<p>You, my friend, are a straightup genius. Thanks so much. I can't believe I spent so much time doing this problem wrong...lol!</p>