L'Hospital's rule is quite alright. But the fastest way to getting at this one is:
let f(x) =

4cos
2x/2+cosx. Then f(

) = 2
So, the problem is lim (f(x) - f(

))/(x-

) which is nothing but f'(x) at x=

.
x


You get the answer in no time as zero.
(x-

) in the denominator when x


gives the 1st clue that some mischief like this is intended.