मराठी

lim x → ∞ √ x 2 − 1 2 x + 1

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प्रश्न

\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\] 

पर्याय

  • 1

  • 0

  • −1 

  • 1/2 

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उत्तर

1/2 

\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
\[ = \lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x^2}}}{2 + \frac{1}{x}}\]
\[\text{ Dividing the numerator and the denominator by x, we get }: \]
\[ = \frac{\sqrt{1 - 0}}{2 + 0}\]
\[ = \frac{1}{2}\]

Hence, the correct answer is d.

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पाठ 29: Limits - Exercise 29.13 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.13 | Q 13 | पृष्ठ ७८

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