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Lim X → ∞ 3 X 3 − 4 X 2 + 6 X − 1 2 X 3 + X 2 − 5 X + 7

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प्रश्न

\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 

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उत्तर

\[\lim_{x \to \infty} \left[ \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7} \right]\]
\[\text{ Dividing the numerator and the denominator by } x^3 : \]
\[ \lim_{x \to \infty} \left[ \frac{3 - \frac{4}{x} + \frac{6}{x^2} - \frac{1}{x^3}}{2 + \frac{1}{x} - \frac{5}{x^2} + \frac{7}{x^3}} \right]\]
\[\text{ When } x \to \infty , \text{ then } \frac{1}{x}, \frac{1}{x^2}, \frac{1}{x^3} \to 0 . \]
\[ \Rightarrow \frac{3}{2}\]

 

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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 2 | पृष्ठ ३८

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