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Lim X → 2 √ 1 + √ 2 + X − √ 3 X − 2 is Equal to - Mathematics

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प्रश्न

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

विकल्प

  • \[\frac{1}{8\sqrt{3}}\]

  • \[\frac{1}{\sqrt{3}}\]

  • $\mathnormal{8 \sqrt{3}}$ 

  • \[\sqrt{3}\]

MCQ
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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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अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ८०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 27 | पृष्ठ ८०

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