Question

Find the value of $x\stackrel{lim}{\to }1\frac{\sqrt{x-1}+\sqrt{x}-1}{\sqrt{x^2-1}}$

Answer 1

We have,

${lim}_{x\to 1}\frac{\sqrt{x-1}+\sqrt{x}-1}{\sqrt{x^2-1}}$

Putting the limits we find it to be in $\frac{0}{0}$ form using L' Hospitals rule.

$={lim}_{x\to 1}\frac{\frac{1}{2\sqrt{x-1}}+\frac{1}{2\sqrt{x}}}{\frac{2x}{2\sqrt{x^2-1}}}={lim}_{x\to 1}\frac{\frac{1}{2\sqrt{x-1}}+\frac{1}{2\sqrt{x}}}{\frac{x}{\sqrt{x^2-1}}}={lim}_{x\to 1}\frac{\frac{\sqrt{x^2-1}}{2\sqrt{x-1}}+\frac{\sqrt{x^2-1}}{2\sqrt{x}}}{x}={lim}_{x\to 1}\frac{\frac{\sqrt{x-1}\sqrt{x+1}}{2\sqrt{x-1}}+\frac{\sqrt{x^2-1}}{2\sqrt{x}}}{x}={lim}_{x\to 1}\frac{\frac{\sqrt{x+1}}{2}+\frac{\sqrt{x^2-1}}{2\sqrt{x}}}{x}$

Putting the limits we get

$=\frac{\frac{\sqrt{2}}{2}}{1}=\frac{1}{\sqrt{2}}$

Hence, this is the answer.