Write the function as:
((2x)/pi +cosx)^(1/cosx) = ( e^ ln((2x)/pi +cosx))^(1/cosx) = e^ (ln((2x)/pi +cosx)/cosx)
Consider now the limit:
lim_(x->pi/2) ln((2x)/pi +cosx)/cosx
which is in the indeterminate form 0/0 and apply l'Hospital's rule:
lim_(x->pi/2) ln((2x)/pi +cosx)/cosx = lim_(x->pi/2) (d/dx ln((2x)/pi +cosx))/(d/dx cosx)
lim_(x->pi/2) ln((2x)/pi +cosx)/cosx = lim_(x->pi/2) -1/sinx (2/pi -sinx)/((2x)/pi+cosx)
lim_(x->pi/2) ln((2x)/pi +cosx)/cosx = (-1) (2/pi-1)/(1+0) = 1-2/pi
As the exponential function is continuous for every x in RR, then:
lim_(x->pi/2) e^ (ln((2x)/pi +cosx)/cosx) = e^( (lim_(x->pi/2) ln((2x)/pi +cosx)/cosx )) = e^(1-2/pi)
