Evaluate the following limit: _ x 1 x^2-1 + x-1 x^2-1 ,x>1

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}},\text{x}>1$

✓

Answer

$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}\times\frac{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}\times\frac{\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^-1}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\big[\big(\text{x}^2-1\big)-(\text{x}-1)\big]\times\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\text{x}\big)\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\sqrt{\text{x}^2-1}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}{(\text{x}+1)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}-1\big)}$
$=\frac{\sqrt{2}}{2\big(\sqrt{2}-1\big)}$
$=\frac{\sqrt{2}}{2\times\big(\sqrt{2}-1\big)}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$
$=\frac{\sqrt{2}+1}{\sqrt{2}}$