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Math Help - Limit indeterminate

  1. #1
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    Limit indeterminate

    Looking for help for:

    Limit as x->infinity

    (sqrt(x)^sqrt(x))/(2^(x^0.6))
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  2. #2
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    Quote Originally Posted by fpver View Post
    Looking for help for:

    Limit as x->infinity

    (sqrt(x)^sqrt(x))/(2^(x^0.6))

    \mathop {\lim }\limits_{x \to \infty } \frac{{{{\sqrt x }^{\sqrt x }}}}<br />
{{{2^{{x^{0.6}}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{\sqrt x /2}}}}<br />
{{{2^{{x^{3/5}}}}}} = \mathop {\lim }\limits_{x \to \infty } \exp \ln \frac{{{x^{\sqrt x /2}}}}<br />
{{{2^{{x^{3/5}}}}}} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \left\{ {\ln {x^{\sqrt x /2}} - \ln {2^{{x^{3/5}}}}} \right\} = \mathop {\lim }\limits_{x \to \infty } \exp \left\{ {\frac{{\sqrt x }}{2}\ln x - {x^{3/5}}\ln 2} \right\} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\left( {\frac{{\sqrt x }}{2}\ln x - {x^{3/5}}\ln 2} \right)\left( {\frac{{\sqrt x }}<br />
{2}\ln x + {x^{3/5}}\ln 2} \right)}}<br />
{{\frac{{\sqrt x }}<br />
{2}\ln x + {x^{3/5}}\ln 2}} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{x}<br />
{4}{{\ln }^2}x - {x^{6/5}}{{\ln }^2}2}}<br />
{{\frac{{\sqrt x }}<br />
{2}\ln x + {x^{3/5}}\ln 2}} = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{1}<br />
{4}{x^{ - 1/5}}{{\ln }^2}x - {{\ln }^2}2}}<br />
{{\frac{1}<br />
{2}{x^{ - 7/10}}\ln x + \frac{{\ln 2}}<br />
{{{x^{3/5}}}}}} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{1}<br />
{4}{{\ln }^2}\exp \frac{{\ln x}}<br />
{{{x^{1/10}}}} - {{\ln }^2}2}}<br />
{{\frac{1}<br />
{2}\ln \exp \frac{{\ln x}}<br />
{{{x^{7/10}}}} + \frac{{\ln 2}}<br />
{{{x^{3/5}}}}}} = {\text{ }}\mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{1}<br />
{4}{{\ln }^2}\exp \frac{{{{\left( {\ln x} \right)}^\prime }}}<br />
{{{{\left( {{x^{1/10}}} \right)}^\prime }}} - {{\ln }^2}2}}<br />
{{\frac{1}<br />
{2}\ln \exp \frac{{{{\left( {\ln x} \right)}^\prime }}}<br />
{{{{\left( {{x^{7/10}}} \right)}^\prime }}} + \frac{{\ln 2}}<br />
{{{x^{3/5}}}}}} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{1}<br />
{4}{{\ln }^2}\exp \frac{{10}}<br />
{{{x^{1/10}}}} - {{\ln }^2}2}}<br />
{{\frac{1}<br />
{2}\ln \exp \frac{{10}}<br />
{{7{x^{7/10}}}} + \frac{{\ln 2}}<br />
{{{x^{3/5}}}}}} = \exp \frac{{\frac{1}<br />
{4}{{\ln }^2}\exp 0 - {{\ln }^2}2}}<br />
{{\frac{1}<br />
{2}\ln \exp 0 + 0}} =

    = \exp \frac{{0 - {{\ln }^2}2}}{{0 + 0}} = \exp \left( { - \infty } \right) = 0.
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