Posted by khela on Thursday, October 28, 2010 at 10:01pm.
can you use L'Hopital's rule?
if so
lim (cos x-1)/ (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 x-1)/x) * x(2sinxcosx) ]
= lim( (cosx - 1)/x) * lim (x/sinx) * lim (1/2cosx)
= 0(1)(1/2) = 0
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