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Lim X → ∞ [ √ X { √ X + 1 − √ X } ]

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प्रश्न

\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\] 

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उत्तर

\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[ = \lim_{x \to \infty} \left[ \sqrt{x}\left\{ \left( \sqrt{x + 1} - \sqrt{x} \right)\frac{\sqrt{x + 1} + \sqrt{x}}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right\} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{x}\left\{ \left( x + 1 \right) - x \right\}}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right]\]

Dividing the numerator and the denominator by\[\sqrt{x}\] 

\[\lim_{x \to \infty} \left[ \frac{1}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \right]\]
\[As x \to \infty , \frac{1}{x} \to 0\]
\[ = \frac{1}{\sqrt{1 + 0} + 1}\]
\[ = \frac{1}{2}\]

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अध्याय 29: Limits - Exercise 29.6 [पृष्ठ ३९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.6 | Q 18 | पृष्ठ ३९

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