We have:
#lim_(x->0)(tan^2(x))/(3x)#
Since we get #0/0# after substituting #0# in the place of #x#, we can use the L'Hopital's Rule, which states that:
If #(f(c))/(g(c))# gives you an indeterminate form, then #lim_(x->c)=(f'(c))/(g'(c))#
We now have:
#lim_(x->0)(tan^2(x))/(3x)=(d/dx[tan^2(x)])/(d/dx[3x])#
A few things to remember:
#d/dx[tan(x)]=sec^2(x)#
#d/dx[x^n]=nx^(n-1)# If #n# is a constant.
#d/dx[f(g(x))]=f'(g(x))*g'(x)#
#=>(2*tan(x)*d/dx[tan(x)])/(3)#
#=>(2*tan(x)*sec^2(x))/(3)#
We now substitute #0# in the place of #x#.
#=>(2*tan(0)*sec^2(0))/(3)=0#
Therefore, #lim_(x->0)(tan^2(x))/(3x)=0#
