Answer
$0$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} \dfrac{1-\ln x}{e^{1/x}}$.
But $\lim\limits_{x \to 0^{+}} \dfrac{1-\ln x}{e^{1/x}}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{A(x) }{B(x)}=\lim\limits_{x \to a}\dfrac{A'(x)}{B'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0^{+}} \dfrac{1-\ln x}{e^{1/x}}=\lim\limits_{x \to 0^{+}} \dfrac{-1/x}{(-1/x^2) (e^{1/x})} \\= \lim\limits_{x \to 0^{+}} \dfrac{e^{-1/x}}{1/x}=\dfrac{\infty}{\infty}$
Again apply L'Hopital's rule :
$ \lim\limits_{x \to 0^{+}} \dfrac{e^{-1/x}}{1/x}= - \lim\limits_{x \to 0^{+}} e^{-1/x}=0$
