#[1]" "lim_(x->0)(cosx)^(1/x^2)#
This is an indeterminate form of the type #1^oo#. You need to first convert it to the form #0/0# or #oo/oo# so you can use L'Hopital's Rule. We can do this by using #e# and #ln#.
#[2]" "=lim_(x->0)e^ln[(cosx)^(1/x^2)]#
#[3]" "=lim_(x->0)e^[(1/x^2)ln(cosx)]=lim_(x->0)e^[ln(cosx)/x^2]#
We can take out #e#.
#[4]" "=e^(lim_(x->0)ln(cosx)/x^2)#
This is now an indeterminate form of the type #0/0#. We can use L'Hopital's Rule now. Get the derivatives of both the numerator and denominator.
#[5]" "=e^(lim_(x->0)(-sinx/cosx)/(2x))=e^(lim_(x->0)(-sinx/(2xcosx))#
This is still indeterminate so you must apply L'Hopital's Rule again.
#[6]" "=e^(lim_(x->0)(-cosx/(2(-xsinx+cosx)))#
You can now get the limit by substitution.
#[6]" "=e^(-cos0/(2(-0sin0+cos0)))#
#[7]" "=e^(-1/2)#
#[8]" "=color(blue)(1/sqrt(e))#
