\displaystyle{\mathop {\lim }\limits_{x \to \infty } {{e^x f\left( {e^x } \right)} \over {f\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } {{8xe^x \left( {\ln 2 + \ln x} \right)^{\sqrt 2 } } \over {8e^x \left( {\ln 2 + \ln e^x } \right)^{\sqrt 2 } }} = \mathop {\lim }\limits_{x \to \infty } x\left( {{{\ln 2 + \ln x} \over {\ln 2 + x}}} \right)^{\sqrt 2 } = 0 < 1}