Question

Evaluate the limit:

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

Answer 1

Given: $\underset{x\to 1}{lim}\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}},x>1$ becomes $\frac{0}{0}$ form

Using Factorization method

$\Rightarrow \underset{x\to 1}{lim}\frac{\sqrt{x2-1^2}+\sqrt{x-1}}{\sqrt{x^2-1^2}}$

$\Rightarrow \underset{x\to 1}{lim}\frac{\sqrt{(x-1)(x+1)}+\sqrt{x-1}}{\sqrt{(x-1)(x+1)}}$

$[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow \underset{x\to 1}{lim}\frac{\sqrt{x-1}(\sqrt{(x+1)}+1)}{\sqrt{(x-1)}\sqrt{(x+1)}}$

$\Rightarrow \underset{x\to 1}{lim}\frac{\sqrt{(x+1)}+1}{\sqrt{(x+1)}}=\frac{\sqrt{(1+1)}+1}{\sqrt{(1+1)}}$

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

$Therefore,$

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