#lim_(xrarr1)(cos((pix)/2))/(cos(1/x)) #
#lim_(xrarr1)(cos((pix)/2))/(cos(1/x)) = (cos(pi/2))/(cos(1)) #
#lim_(xrarr1)(cos((pix)/2))/(cos(1/x)) = 0/cos(1) = "undefined" or oo#
I wont describe cos(1) because finding it would be useless. It would be approx 0.5
So apply L'Hopital's rule
Numerator
#d/dx (cos((pix)/2)) = d/dxcos((pix)/2)d/dx(pix)/2#
#-(pi)/2 sin((pix)/2)#
Plug in 1
#= -(pi)/2 sin((pixx1)/2)#
# = -pi/2#
Denominator
# d/dx(cos(1/x)) = d/dx(cos(1/x))d/dx(1/x)#
#= d/dx(cos(1/x))d/dx (x^(-1))#
#= (-sin(1/x)) (-1xx x^(-1 xx -1))#
#= (-sin(1/x)) (-x^(-2))#
#= (cancel(-)sin(1/x)) (cancel(-)1/(x^(2)))#
#= sin(1/x)/x^2#
Plug in 1
#= sin(1)/1^2#
#= 0.8414709848#
#"If we consider " 0.8414709848 ~~ 1" then "#
#lim_(xrarr1)(cos((pix)/2))/(cos(1/x)) = (-pi)/2#
But I'm not approximating so the answer may be not be there if it is a choice question
Divide numerator by denominator
#= (-pi/2)/0.8414709848#
# = -1.86672666694 #
#(-pi)/2 ~~-1.57079632679~~-1.86672666694 #
#therefore lim_(xrarr1)(cos((pix)/2))/(cos(1/x)) != (pi)/2 #
