Evaluate the following limits: _ x 2 [ 1+2+x -3 x-2 ] - Vidyadip

Question

Evaluate the following limits:
$\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right]$

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Answer

$ \lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$
$=\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} \times \frac{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}\right] $
...[By rationalization]
$ =\lim _{x \rightarrow 2} \frac{1+\sqrt{2+x}-3}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})}$
$=\lim _{x \rightarrow 2} \frac{\sqrt{2+x}-2}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})} \times \frac{\sqrt{2+x}+2}{\sqrt{2+x}+2} $
...[By rationalization]
$ =\lim _{x \rightarrow 2} \frac{2+x-4}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\lim _{x \rightarrow 2} \frac{x-2}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\lim _{x \rightarrow 2} \frac{1}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\frac{1}{(\sqrt{1+\sqrt{2+2}}+\sqrt{3})(\sqrt{2+2}+2)}$
$\left.=\frac{1}{(\sqrt{1+2}+\sqrt{3})(2+2)} \quad \therefore x-2 \neq 0\right]$
$=\frac{1}{(2 \sqrt{3})(4)}$
$=\frac{1}{8 \sqrt{3}} $

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