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Lim X → ∞ 3 X − 1 + 4 X − 2 5 X − 1 + 6 X − 2

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प्रश्न

\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]

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उत्तर

\[\lim_{x \to \infty} \left[ \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\frac{3}{x} + \frac{4}{x^2}}{\frac{5}{x} + \frac{6}{x^2}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{3x + 4}{5x + 6} \right]\]
\[\text{ Dividingthe numerator and the denominator by } x:\]

\[ \lim_{x \to \infty} \left[ \frac{3 + \frac{4}{x}}{5 + \frac{6}{x}} \right] = \frac{3}{5}\]

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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 9 | पृष्ठ ३९

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