First slide

Introduction to limits

Question

$lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}$ is equal to

Moderate

A

$\frac{1}{8 \sqrt{3}}$
B

$\frac{1}{\sqrt{3}}$
C

$8 \sqrt{3}$
D

$\sqrt{3}$

Solution

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

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