Question

Find the following limits.
$\underset{x\to 1}{lim}\frac{\sqrt{x}-1}{x^2-1}.$

Answer 1

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

By using factorization

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

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

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

$\underset{x\to 1}{lim}\frac{1}{(\sqrt{1}+1)(1+1)}=\frac{1}{2\times 2}=\frac{1}{4}$.