The value oflimx→2 x3−4xx3−8−1−x+2xx−2−2x−2−1, is

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

1. A

$\frac{1}{2}$

2. B

2

3. C

1

4. D

none of these

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### Solution:

We have,

$\begin{array}{l}\underset{x\to 2}{lim} \left\{{\left(\frac{{x}^{3}-4x}{{x}^{3}-8}\right)}^{-1}-{\left(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)}^{-1}\right\}\\ =\underset{x\to 2}{lim} \left\{\frac{{x}^{2}+2x+4}{x\left(x+2\right)}-{\left(\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{2}\right)}{x-2}-\frac{\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)}{x-2}\right)}^{-1}\right\}\\ =\underset{x\to 2}{lim} \left\{\frac{{x}^{2}+2x+4}{x\left(x+2\right)}-{\left(\frac{\left(\sqrt{x}+\sqrt{2}\right)\left(\sqrt{x}-\sqrt{2}\right)}{x-2}\right)}^{-1}\right\}\\ =\underset{x\to 2}{lim} \left\{\frac{{x}^{2}+2x+4}{x\left(x+2\right)}-1\right\}=\frac{12}{8}-1=\frac{1}{2}\end{array}$

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