Question

The value of $\underset{x\to 2}{lim}({\left(\frac{x^3-4x}{x^3-8}\right)}^{-1}-{(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}})}^{-1})$ is equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $\frac{1}{2}$

$\underset{x\to 2}{lim}({\left(\frac{x^3-4x}{x^3-8}\right)}^{-1}-{(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}})}^{-1})$
$=\underset{x\to 2}{lim}({\left(\frac{x(x^2-4)}{x^3-8}\right)}^{-1}-{(\frac{\sqrt{x}(\sqrt{x}+\sqrt{2})}{(\sqrt{x}+\sqrt{2})(\sqrt{x}-\sqrt{2})}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}})}^{-1})$

$=\underset{x\to 2}{lim}({\left(\frac{x(x-2)(x+2)}{(x-2)(x^2+4+2x)}\right)}^{-1}-{\left(\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)}^{-1})$

$=\underset{x\to 2}{lim}({\left(\frac{x(x+2)}{(x^2+4+2x)}\right)}^{-1}-1)$
$=\frac{3}{2}-1=\frac{1}{2}$​​​​