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प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
विकल्प
\[\frac{1}{8\sqrt{3}}\]
\[\frac{1}{\sqrt{3}}\]
$\mathnormal{8 \sqrt{3}}$
\[\sqrt{3}\]
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उत्तर
\[(a)\]
\[ \because \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} = \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} \times \frac{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}}{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}}\]
\[ = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)}\]
\[ = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)} \times \frac{\sqrt{2 + x} + 2}{\sqrt{2 + x} + 2}\]
\[ = \lim_{x \to 2} \frac{\left( x - 2 \right)}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)\left( \sqrt{2 + x} + 2 \right)}\]
\[ = \lim_{x \to 2} \frac{1}{\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)\left( \sqrt{2 + x} + 2 \right)}\]
\[ = \frac{1}{\left( \sqrt{1 + \sqrt{2 + 2}} + \sqrt{3} \right)\left( \sqrt{2 + 2} + 2 \right)}\]
\[ = \frac{1}{2\sqrt{3} \times 4}\]
\[ = \frac{1}{8\sqrt{3}}\]
\[\]
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