First slide

Introduction to limits

Question

The value of
$lim_{x \to 2} \{{(\frac{x^{3} - 4 x}{x^{3} - 8})}^{- 1} - {(\frac{x + \sqrt{2 x}}{x - 2} - \frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}})}^{- 1}\}$ , is

Moderate

A

$\frac{1}{2}$
B

2
C

1
D

none of these

Solution

We have,
$\begin{array}{l} lim_{x \to 2} \{{(\frac{x^{3} - 4 x}{x^{3} - 8})}^{- 1} - {(\frac{x + \sqrt{2 x}}{x - 2} - \frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}})}^{- 1}\} \\ = lim_{x \to 2} \{\frac{x^{2} + 2 x + 4}{x (x + 2)} - {(\frac{\sqrt{x} (\sqrt{x} + \sqrt{2})}{x - 2} - \frac{\sqrt{2} (\sqrt{x} + \sqrt{2})}{x - 2})}^{- 1}\} \\ = lim_{x \to 2} \{\frac{x^{2} + 2 x + 4}{x (x + 2)} - {(\frac{(\sqrt{x} + \sqrt{2}) (\sqrt{x} - \sqrt{2})}{x - 2})}^{- 1}\} \\ = lim_{x \to 2} \{\frac{x^{2} + 2 x + 4}{x (x + 2)} - 1\} = \frac{12}{8} - 1 = \frac{1}{2} \end{array}$

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