Let's split up the limit up.
#lim_(x->0)(root(3)(1+x^2)-root(4)(1-2x))/(x+x^2)=lim_(x->0)(root(3)(1+x^2))/(x+x^2)-lim_(x->0)(root(4)(1-2x))/(x+x^2)#
Convert the roots into exponents. Recall that #root(a)x=x^(1/a)#
#lim_(x->0)(root(3)(1+x^2))/(x+x^2)-lim_(x->0)(root(4)(1-2x))/(x+x^2)=lim_(x->0)(1+x^2)^(1/3)/(x+x^2)-lim_(x->0)(1-2x)^(1/4)/(x+x^2)#
For #lim_(x->0)(1+x^2)^(1/3)/(x+x^2)#, let's multiply by #(1+x^2)^(2/3)/(1+x^2)^(2/3).#
This will get the root out of the numerator and allow us to get rid of the indeterminate form #0/0# that we would otherwise end up with if we evaluated right away. Really, it's just the same as multiplying by #1.#
#lim_(x->0)(1+x^2)^(1/3)/(x+x^2)*(1+x^2)^(2/3)/(1+x^2)^(2/3)=lim_(x->0)(1+x^2)/((1+x^2)^(2/3)(x+x^2))#
Evaluating, we get
#lim_(x->0)(1+x^2)/((1+x^2)^(2/3)(x+x^2))=(1+0^2)/((1+0^2)^(2/3)(0+0^2))=1/1=1#
For #lim_(x->0)(1-2x)^(1/4)/(x+x^2)#, let's multiply by #(1-2x)^(3/4)/(1-2x)^(3/4).# This will get rid of the root in the numerator, and is also just really multiplying by #1# :
#lim_(x->0)(1-2x)^(1/4)/(x+x^2)*(1-2x)^(3/4)/(1-2x)^(3/4)=lim_(x->0)(1-2x)/((1-2x)^(3/4)(x-x^2))#
Evaluating, we get
#lim_(x->0)(1-2x)/((1-2x)^(3/4)(x-x^2))=(1-2*0)/((1-2*0)^(3/4)(0-0^2))=1/(1)^(3/4)=1#
So, now that we have both of our limits, we can solve:
#lim_(x->0)(root(3)(1+x^2))/(x+x^2)-lim_(x->0)(root(4)(1-2x))/(x+x^2)=1-1=0#
