First slide

Introduction to limits

Question

The value of $lim_{x \to 2} \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}$ is

Moderate

A

1
B

2
C

$\frac{1}{2}$
D

$\frac{1}{4}$

Solution

$lim_{x \to 2} \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}$
At x $\to$ 2,it is an indeterminate $(\frac{0}{0})$ form.
On applying Li Hospital's rule, we get
$\begin{matrix} lim_{x \to 2} \frac{\frac{1}{2 \sqrt{x - 2}} \cdot 1 + \frac{1}{2 \sqrt{x}} - 0}{\frac{1}{2 \sqrt{x^{2} - 4}} \cdot 2 x} = lim_{x \to 2} \frac{\frac{1}{2} [\frac{1}{\sqrt{x - 2}} + \frac{1}{\sqrt{x}}]}{\frac{x}{\sqrt{x - 2} \sqrt{x + 2}}} \\ = lim_{x \to 2} \frac{\frac{1}{2} [\frac{\sqrt{x} + \sqrt{x - 2}}{\sqrt{x} \sqrt{x - 2}}]}{\frac{x}{\sqrt{x - 2} \sqrt{x + 2}}} \\ = lim_{x \to 2} \frac{1}{2} [\frac{\sqrt{x} + \sqrt{x - 2}}{x \sqrt{x}}] \sqrt{x + 2} \\ = \frac{1}{2} [\frac{\sqrt{2}}{2 \sqrt{2}}] \sqrt{2 + 2} \\ = \frac{1}{2} \cdot [\frac{1}{2}] \cdot 2 = \frac{1}{2} \end{matrix}$

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