1. Limit indeterminate

Looking for help for:

Limit as x->infinity

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

2. Originally Posted by fpver
Looking for help for:

Limit as x->infinity

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

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

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

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

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

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

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

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