## Calculus (3rd Edition)

$\frac{1}{2}$
We know that $\lim\limits_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$ Then, $\lim\limits_{x \to 1}\frac{\sqrt x-1}{x-1}=\lim\limits_{x \to 1}\frac{x^{1/2}-1^{1/2}}{x-1}=\frac{1}{2}×1^{\frac{1}{2}-1}$ =$\frac{1}{2}×\frac{1}{\sqrt 1}= \frac{1}{2}$