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December 22, 2014

December 22, 2014

Posted by **khela** on Thursday, October 28, 2010 at 10:01pm.

x->0

- maths -
**Reiny**, Thursday, October 28, 2010 at 10:10pmcan 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

- maths -
**khela**, Thursday, October 28, 2010 at 10:11pmno u cant

- maths -
**Reiny**, Thursday, October 28, 2010 at 10:20pmsome 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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