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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Thursday, June 28, 2012 at 8:47am.

- calculus -
**Reiny**, Thursday, June 28, 2012 at 10:24amUsually the first step I take in doing a limit question is to actually substitute the approach value, so I got

lim cos 4x/(1- cos 2x) , as x -->0

--> 1/(1-0)

which is undefined,

so there is not limit, or the result is infinity

check: I tried x = .0001

and got 52631574.74

- calculus -
**Steve**, Thursday, June 28, 2012 at 11:18amhowever, a more careful reading reveals :-)

lim (1-cos4x)/(1-cos2x)

= lim 4sin4x/2sin2x

= lim (4sin2x cos2x)/sin2x

= lim 4cos2x

= 4

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