As the function is continuous at every point except #x=0#, we can solve the first one by direct substitution:
#lim_(x->pi/2)(2x-sin(2x))/x^2 =(2(pi/2)-sin(2(pi/2)))/(pi/2)^2#
#=(pi-sin(pi))/(pi^2/4)#
#=(4(pi-0))/pi^2#
#=4/pi#
For the second one, direct substitution results in a #0/0# indeterminate form, so we will apply L'Hopital's rule.
#lim_(x->0)(2x-sin(2x))/x^2 = lim_(x->0)(d/dx(2x-sin(2x)))/(d/dxx^2)#
#=lim_(x->0)(2-2cos(2x))/(2x)#
#=lim_(x->0)(1-cos(2x))/x#
Direct substitution still results in a #0/0# indeterminate form, meaning we can apply L'Hopital's rule once again.
#lim_(x->0)(1-cos(2x))/x = lim_(x->0)(d/dx(1-cos(2x)))/(d/dxx)#
#=lim_(x->0)(2sin(2x))/1#
#=lim_(x->0)2sin(2x)#
As the new function is continuous at #0#, we may now use direct substitution.
#lim_(x->0)2sin(2x) = 2sin(2*0) = 2sin(0) = 0#
Thus #lim_(x->0)(2x-sin(2x))/x^2 = 0#
