#Lim_(x->0)((8+h)^(1/3)-2)/h#
Now notice that we need to get rid of exponent #1/3#. Therefore we multiply the numerator and denominator by an expression that will make it free of exponent #1/3#.
We know that: #A^3-B^3=(A-B)(A^2+AB+B^2)#
In our case #A=(8+h)^(1/3)# and #B=2#. I notice that if I multiply numerator and denominator by #(A^2+AB+B^2)# i.e., #((8+h)^(2/3)+2(8+h)^(1/3)+4)#, then I will get #A^3=((8+h)^(1/3))^3=(8+h)#. and hence A becomes free of exponent #1/3#.
Multiply and divide by #((8+h)^(2/3)+2(8+h)^(1/3)+4)#
#=>Lim_(x->0)(((8+h)^(1/3)-2)/h*((8+h)^(2/3)+2(8+h)^(1/3)+4)/((8+h)^(2/3)+2(8+h)^(1/3)+4))#
#=>Lim_(x->0)((8+h)^(3/3)-2^3)/(h*((8+h)^(2/3)+2(8+h)^(1/3)+4))#
#=>Lim_(x->0)(8+h-8)/(h*((8+h)^(2/3)+2(8+h)^(1/3)+4))#
#=>Lim_(x->0)(h)/(h*((8+h)^(2/3)+2(8+h)^(1/3)+4))#
#=>Lim_(x->0)(color(red)cancel(h))/(color(red)cancel(h)*((8+h)^(2/3)+2(8+h)^(1/3)+4))#
#=>Lim_(x->0)1/((8+h)^(2/3)+2(8+h)^(1/3)+4)#
#=>1/((8+0)^(2/3)+2(8+0)^(1/3)+4)#
#=>1/(4+2*2+4)#
#=>1/12#
