Quote (lordkira @ Wed, May 16 2007, 09:32pm)
Quote (safmy @ Wed, May 16 2007, 03:36pm)
Paper three complete. Easy question for you guys
Find the limit of (cos x - 1)/ (x sin x) as n
0
Free fg for those who can (this is actually the easier questions of the paper, i.e 1. b )
Oh ok
I will assume that n is supposed to be an 'x' and that it was a typo
let's me try (it's been a while but bear with me ya'll)
I will exclude the lim of x-->0 part until I am substituting the value 0 for x
(cos x - 1) / (x sin x)
(-sin x) / (sin x - x cos x)
(-cos x) / (cos x + cos x - x sin x)
(-cos x) / (2cos x - x sin x)
lim of (-cos x) / (2cos x - x sin x) as x-->0 is -1/2You made a small mistake with the product rule but you still got the correct end result somehow u crazy man. (then again I did make the mistake of putting n
0 when it's x)
I'll do the functions with some explanations to enlighten the guildies (unlike some evil geniuses)
Since lim x
for f(x)/g(x) gives us 0/0 we have to use the l'Hopital's rule to find the limit (that is we derive each function separately), giving:-
f'(x)/g'(x)= (-sin x) / (x cos x + sin x)
finding the derivative again since the limit of this function gives 0/0 still
= (-cos x) / (-x sin x + cos x + cos x)
= (-cos x) / (2cos x - x sin x )
therefore lim x--> 0 for f(x)/g(x) = -1/2
Present for kira will arrive soon (even though he cheated)