Dear friends. I would like to enlist the help of you in this question. I tried to improve the algebric expression, I tried the L'Hospital theorem, but nothing worked.
Grateful for your help. Big hug to everyone.
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Hint: Let $x=y^6$. We are interested in the limit as $y\to 1$ of $$\frac{1}{2(1-y^3)}-\frac{1}{3(1-y^2)}.$$ Bring to the common denominator $(2)(3)(1-y^2)(1-y^3)$. The numerator is $3(1-y^2)-2(1-y^3)$. There is an obvious factor of $1-y$ on top. Simplify. There is an additional slightly hidden factor of $1-y$.
André Nicolas
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Dear André Nicolas, is the answer 1/12? – Curso FDX Mar 03 '14 at 00:41
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Yes. After the simplification we have $\frac{1+2y}{6(1+y)(1+y+y^2)}$. – André Nicolas Mar 03 '14 at 01:50