## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)=\infty$$
To solve this limit follow the steps below: $$\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)=\lim_{x\to1^+}\left(\frac{1}{x-1}-\frac{\sqrt{x-1}}{x-1}\right)=\lim_{x\to1^+}\frac{1-\sqrt{x-1}}{x-1}=\left[\frac{1-\sqrt{1^+-1}}{1^+-1}\right]=\left[\frac{1-0^+}{0^+}\right]=\infty.$$