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Lim X → ∞ √ X 2 + 7 X − X

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प्रश्न

\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 

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उत्तर

\[\lim_{x \to \infty} \left[ \sqrt{x^2 + 7x} - x \right]\]
\[\text{ Rationalising the numerator }: \]
\[ \lim_{x \to \infty} \left[ \left( \sqrt{x^2 + 7x} - x \right) \frac{\left( \sqrt{x^2 + 7x} + x \right)}{\left( \sqrt{x^2 + 7x} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x^2 + 7x - x^2}{\left( \sqrt{x^2 + 7x} + x \right)} \right]\]
\[\text{ Dividing the numerator and the denominator by } x: \]
\[ \lim_{x \to \infty} \left[ \frac{7}{\frac{\sqrt{x^2 + 7x}}{x} + 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{7}{\frac{\sqrt{x^2 + 7x}}{x} + 1} \right]\]
\[\text{ When }x \to \infty , \text{ then } \frac{1}{x} \to 0 . \]
\[ \Rightarrow \frac{7}{\sqrt{1} + 1}\]
\[ = \frac{7}{2}\]

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अध्याय 29: Limits - Exercise 29.6 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.6 | Q 6 | पृष्ठ ३८

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