Chapter 3 - The Derivative In Graphing And Applications - 3.6 Rectilinear Motion - Exercises Set 3.6 - Page 245: 16

$\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} =0 $

$\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{ \frac{d(1-\ln x)}{dx}}{ \frac{d(e^{ \frac{1}{x}})}{dx}} $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{- \frac{1}{x}}{ e^{\frac{1}{x}}\frac{d}{dx} ( \frac{1}{x}) } $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{- \frac{1}{x}}{ e^{\frac{1}{x}}\frac{d}{dx} ( x^{-1} )} $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{- \frac{1}{x}}{ e^{\frac{1}{x}} (-x^{-2})} $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{(x^2) \frac{1}{x}}{ e^{\frac{1}{x}} } $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{x}{ e^{\frac{1}{x}} } $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{\frac{dx}{dx}}{ \frac{d(e^{ \frac{1}{x}})}{dx}} $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = \lim\limits_{x \to 0^{+}} \frac{1}{-1x^{-2}} $ $\lim\limits_{x \to 0^{+}} \frac{1- \ln x}{ e^{\frac{1}{x}}} = - \lim\limits_{x \to 0^{+}} x^2=0 $