Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$

Answer

$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{1}\big)}$
$=\frac{4}{2(1+1)}$
$=\frac{4}{4}=1$

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