Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 322: 84

$2$

Here, we have $\lim\limits_{x \to \infty}\dfrac{1}{\sqrt x}\int_1^{x} \dfrac{dt}{\sqrt t}=\lim\limits_{x \to \infty} (\dfrac{1}{\sqrt x}) |2\sqrt t|_1^x$ $\lim\limits_{x \to \infty}(\dfrac{1}{\sqrt x}) |2\sqrt t|_1^x=(\dfrac{1}{\sqrt x})\lim\limits_{x \to \infty} 2 \sqrt x-2$ Thus, $\lim\limits_{x \to \infty}(2-\dfrac{2}{\sqrt x})=2-0=2$