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$\lim\limits_{x\to \infty}\big(\large\frac{x+1}{x+2}\big)^{2x+1}$ is

$(a)\;e\qquad(b)\;e^{-2}\qquad(c)\;e^{-1}\qquad(d)\;1$

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$\lim\limits_{x\to \infty}\big(\large\frac{x+1}{x+2}\big)^{2x+1}=$$\lim\limits_{e^{\large x\to\infty}}\big(\large\frac{x+1}{x+2}\normalsize -1\big)(2x+1)$
$\Rightarrow \lim\limits_{e^{\large x\to\infty}}\large\frac{-(2x+1)}{x+2}$
$\Rightarrow e^{-2}$
Hence (b) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 
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