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Lim N → ∞ N 2 1 + 2 + 3 + . . . + N - Mathematics

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प्रश्न

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

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उत्तर

\[\lim_{n \to \infty} \left[ \frac{n^2}{1 + 2 + 3 . . . . . n} \right]\]
\[\text{ It is of the form } \frac{\infty}{\infty} . \]
\[ \Rightarrow \lim_{n \to \infty} \left[ \frac{n^2}{n\frac{\left( n + 1 \right)}{2}} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{2n}{n + 1} \right]\]
\[\text{ Dividing the numerator and the denominator by } n:\]
\[ \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n}}\]
\[ = 2\] 

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अध्याय 29: Limits - Exercise 29.6 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.6 | Q 8 | पृष्ठ ३८

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