We have
#1/x^2-1/sin^2x=(sin^2x-x^2)/(x^2sin^2x)=((sin x/x)^2-1)/(1-cos^2x)=#
#=((sinx/x+1)/(1+cosx))((sinx/x-1)/(1-cosx))#
Now following with #lim_(x->0)((sinx/x-1)/(1-cosx))#
At this point we introduce the series representation for #sinx# and #cos x#
#sinx = x-x^3/(3!)+x^5/(5!)+cdots#
#cosx = 1-x^2/(2!)+x^4/(4!)+cdots#
so
#sinx/x = 1-x^2/(3!)+x^4/(5!)+cdots#
making substitutions
#(sinx/x-1)/(1-cosx)=(-x^2/(!3)+x^4/(5!)-x^6/(7!)+cdots)/(x^2/(2!)-x^4/(4!)+x^6/(6!)+cdots)=(-1/(!3)+x^2/(5!)-x^4/(7!)+cdots)/(1/(2!)-x^2/(4!)+x^4/(6!)+cdots)#
then
#lim_(x->0)(sinx/x-1)/(1-cosx)=-1/3# and
#lim_(x->0)(sinx/x+1)/(1+cosx)= 1#
Finally
#lim_(x->0)((sinx/x+1)/(1+cosx))((sinx/x-1)/(1-cosx))=(1)(-1/3)=-1/3#
