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Math Help - A fiendish limit

  1. #1
    Junior Member gusztav's Avatar
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    A fiendish limit

    After struggling with this limit for a while, I would really, really appreciate your help with solving this problem.

    Let a, b, c \in \mathbb{R}_+ be positive real numbers. Without using the l'H˘pital's rule, find the limit

    \lim_{x\to 0}<br />
\left( \frac{a^{x^3}+b^{x^3}+c^{x^3}}<br />
{a^{x^2}+b^{x^2}+c^{x^2}} \right) ^\frac{1}{x^2}


    I tried every method of attack known to me, but to no avail; substitution; trying to utilize the identity \lim_{x\to 0} (1+x)^\frac{1}{x}=e, ... nothing seemed to work.

    Mathematica says the limit should be \frac{1}{\sqrt[3]{abc}}, but I have no idea how to arrive at that.

    I'd be very grateful for any help.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by gusztav View Post
    After struggling with this limit for a while, I would really, really appreciate your help with solving this problem.

    Let a, b, c \in \mathbb{R}_+ be positive real numbers. Without using the l'H˘pital's rule, find the limit

    \lim_{x\to 0}<br />
\left( \frac{a^{x^3}+b^{x^3}+c^{x^3}}<br />
{a^{x^2}+b^{x^2}+c^{x^2}} \right) ^\frac{1}{x^2}


    I tried every method of attack known to me, but to no avail; substitution; trying to utilize the identity \lim_{x\to 0} (1+x)^\frac{1}{x}=e, ... nothing seemed to work.

    Mathematica says the limit should be \frac{1}{\sqrt[3]{abc}}, but I have no idea how to arrive at that.

    I'd be very grateful for any help.
    In the first instance, the special case a = b = c is simple enough and the answer agrees with Mathematica.
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