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प्रश्न
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
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उत्तर
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( n + 2 \right) \left( n + 1 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right) \left( n + 1 \right)! - \left( n + 1 \right)!} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( n + 1 \right)!}{\left( n + 1 \right)!} \times \frac{\left( n + 2 + 1 \right)}{\left( n + 2 - 1 \right)} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n + 3}{n + 1} \right]\]
Dividing the numerator and the denominator by n:
\[\lim_{n \to \infty} \left[ \frac{1 + \frac{3}{n}}{1 + \frac{1}{n}} \right] \]
\[\text{ When n } \to \infty , \text{ then } \frac{1}{n} \to 0 . \]
\[ \Rightarrow \frac{1}{1} = 1\]
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