limn→∞ nn+1α+sin1nn when α ∈ Q is equal to

1∞ form L = elimn→∞ n nn+1α+sin1n-1 = e limn→∞ n sin 1n+ limn→∞ nnn+1α-1 Consider limn→∞ nnn+1α-1 = limn→∞ n11+1/nα-1 Put n=1y . The limy→0 1y11+yα-1= limy→0 1-1+yαy=-α∴ L= e1-α

limn→∞ nn+1α+sin1nn when α ∈ Q is equal to

1∞ form L = elimn→∞ n nn+1α+sin1n-1 = e limn→∞ n sin 1n+ limn→∞ nnn+1α-1 Consider limn→∞ nnn+1α-1 = limn→∞ n11+1/nα-1 Put n=1y . The limy→0 1y11+yα-1= limy→0 1-1+yαy=-α∴ L= e1-α

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$\lim_{n \to \infty} {\{{(\frac{n}{n + 1})}^{α} + \sin \frac{1}{n}\}}^{n} (when α \in Q)$ is equal to

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$e^{- α}$

$- α$

$e^{1 - α}$

$e^{1 + α}$

answer is C.

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Detailed Solution

$\begin{array}{l} 1^{\infty} form \\ L = e^{\lim_{n \to \infty} n \{{(\frac{n}{n + 1})}^{α} + \sin \frac{1}{n} - 1\}} = e_{}^{\lim_{n \to \infty} n \sin \frac{1}{n} + \lim_{n \to \infty} n ({(\frac{n}{n + 1})}^{α} - 1)} \\ Consider \lim_{n \to \infty} n ({(\frac{n}{n + 1})}^{α} - 1) = \lim_{n \to \infty} n ({(\frac{1}{1 + 1 / n})}^{α} - 1) \\ Put n = \frac{1}{y} . The \\ \lim_{y \to 0} \frac{1}{y} ({(\frac{1}{1 + y})}^{α} - 1) = \lim_{y \to 0} \frac{1 - {(1 + y)}^{α}}{y} = - α \\ ∴ L = e^{1 - α} \end{array}$