Math
posted by Alex .
Find the limit of
(2cosx+3cosx2)/(2cosx1)as x tends to pi/3.

if you let y=cosx, then you have
(2u1)(u+2)/(2u1)
so, except when u=1/2, you just have (u+2)
so, the limit is cos pi/3 + 2 = 5/2
there is a removable hole at x=pi/3
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