I haven't tried it yet, but the "traditional" way to handle a limit with a cube root like this is to make use of the fact that [MATH]a^3 - b^3 = (a - b)(a^2 + ab + b^2)[/MATH]. Do you see how this might be used in a similar way to conjugates with square roots?How to find the limit without L'hopital : lim x->0 ( (1+2X) - (1 + 3 x)^(2/3))/(x^2)
That numerator becomes "user-friendly" if you would apply the binomial theorem to the respective portion of the numerator:How to find the limit without L'hopital : lim x->0 ( (1+2X) - (1 + 3 x)^(2/3))/(x^2)