#lim_(x->0) (tanx-x)/x^3#
Substitution give #(tan0-0)/0 rarr 0/0 # this is indeterminate so we can use #color(blue)"L'Hopital's Rule"#
#lim_(x->0) (tanx-x)/x^3=lim_(x->0) (color(blue)(d/dx)*tanx-x)/(color(blue)(d/dx)x^3#
#lim_(x->0) (sec^2x-1)/(3x^2) rarr (sec^2(0)-1)/(3(0)^2#
#(1-1)/(0) rarr 0/0# Indeterminate
Once more use #color(blue)"L'Hopital's Rule"#
#lim_(x->0) (sec^2x-1)/(3x^2)=lim_(x->0)(color(blue)(d/dx)*sec^2x-1)/(color(blue)(d/dx)*3x^2) rarr lim_(x->0)(2sec^2xtanx)/(6x)#
Substitution
#(2sec^2(0)tan(0))/(6(0)) rarr (2*1*0)/(0) rarr 0/0#
You know what's next #color(blue)"L'Hopital's Rule"#
#lim_(x->0)(2sec^2xtanx)/(6x)=lim_(x->0)(color(blue)(d/dx)*2sec^2xtanx)/(color(blue)(d/dx)*6x)#
#lim_(x->0)((2sec^2x)(2tan^2x+sec^2x))/6#
Substitution
#((2sec^2(0))(2tan^2(0)+sec^2(0)))/6 rarr ((2*1)(2*0+1))/6 rarr (2*1)/6#
#:.1/3#
