maths
posted by khela on .
find the lim (cos x1)/ (sin 2x)
x>0

can you use L'Hopital's rule?
if so
lim (cos x1)/ (sin 2x)
x>0
= lim (sinx)/(2cos 2x) as x > 0
= 0/2 = 0 
no u cant

some common trig limits you should have come across:
lim( (sinx)/x ) as x > 0 = 1
lim( x/sinx ) as x > 0 = 1
lim( (cosx  1)/x) as x  0 = 0
so lets rewrite the question as
lim[ ((cos x1)/x) * x(2sinxcosx) ]
= lim( (cosx  1)/x) * lim (x/sinx) * lim (1/2cosx)
= 0(1)(1/2) = 0