Question

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

• A. $\frac{1}{2}$
• B. $2$
• C. $1$
• D. None of these

Answer 1

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

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

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

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

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

$=\frac{4}{2^2+2\cdot 2}=\frac{4}{8}=\frac{1}{2}$