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प्रश्न
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
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उत्तर
1
\[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\]
\[\text{ Dividing } N^r \text{ & } D^r \text{ by } \left( n + 1 \right)!: \]
\[ \Rightarrow \lim_{n \to \infty} \frac{\frac{\left( n + 2 \right) \left( n + 1 \right)!}{\left( n + 1 \right)!} + 1}{\frac{\left( n + 2 \right) \left( n + 1 \right)!}{\left( n + 1 \right)!} - 1}\]
\[ = \lim_{n \to \infty} \frac{n + 2 + 1}{n + 2 - 1}\]
\[ = \lim_{n \to \infty} \frac{n + 3}{n + 1}\]
\[ = \lim_{n \to \infty} \frac{1 + \frac{3}{n}}{1 + \frac{1}{n}}\]
\[ = 1\]
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