- Thread starter yalam
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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 \(\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). 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 theHow to find the limit without L'hopital : lim x->0 ( (1+2X) - (1 + 3 x)^(2/3))/(x^2)

\(\displaystyle (1 + 3x)^{2/3} \ = \ 1^{2/3} \ + \ (\tfrac{1}{1!})(\tfrac{2}{3})(1)^{-1/3}(3x) \ + \ (\tfrac{1}{2!})(\tfrac{2}{3})( \tfrac{-1}{3})(1)^{-4/3}(3x)^2 \ + \ [only \ \ terms \ \ with \ \ degree \ \ higher \ \ than \ \ two] \ = \)

\(\displaystyle 1 \ + \ 2x \ - x^2 \ + \ [only \ \ terms \ \ with \ \ degree \ \ higher \ \ than \ \ two] \)