Question

$\underset{x\to 1}{\lim }\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}},x>1$

Answer 1

$$\begin{gathered} \underset{x\to 1}{\lim }\left[\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to 1}{\lim }\left[\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}+\frac{\sqrt{x-1}}{\sqrt{x^2-1}}\right] \\ =\underset{x\to 1}{\lim }\left[1+\frac{\sqrt{x-1}}{\left(\sqrt{x-1}\right)\left(\sqrt{x+1}\right)}\right] \\ =1+\frac{1}{\sqrt{1+1}} \\ =1+\frac{1}{\sqrt{2}} \\ =\frac{\sqrt{2}+1}{\sqrt{2}} \end{gathered}$$