# Limit indeterminate

• Sep 28th 2009, 12:34 PM
fpver
Limit indeterminate
Looking for help for:

Limit as x->infinity

(sqrt(x)^sqrt(x))/(2^(x^0.6))
• Sep 28th 2009, 02:52 PM
DeMath
Quote:

Originally Posted by fpver
Looking for help for:

Limit as x->infinity

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

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

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

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

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

$\displaystyle = \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}}}}}} =$$\displaystyle {\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}}}}}} =$

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

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