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Lim N → ∞ [ 1 3 + 2 3 + . . . . N 3 N 4 ]

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प्रश्न

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]

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उत्तर

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . + n^3}{n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left[ \frac{n\left( n + 1 \right)}{2} \right]^2}{n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( n + 1 \right)^2}{4 n^2} \right]\]
\[ = \lim_{n \to \infty} \left[ \left( \frac{n + 1}{n} \right)^2 \times \frac{1}{4} \right]\]
\[ = \lim_{n \to \infty} \left[ \left( 1 + \frac{1}{n} \right)^2 \times \frac{1}{4} \right]\]
\[\text{ When } n \to \infty , \text{ then } \frac{1}{n} \to 0 . \]
\[ \Rightarrow \frac{1}{4}\]

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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 16 | पृष्ठ ३९

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