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For $x\in R\lim\limits_{x\to \infty}\big(\large\frac{x-3}{x+2}\big)^x$=

$(a)\;e\qquad(b)\;e^{-1}\qquad(c)\;e^{-5}\qquad(d)\;e^5$

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For $x\in R$
$\lim\limits_{x\to \infty}\big(\large\frac{x-3}{x+2}\big)^x=$$\lim\limits_{x\to \infty}\big[1-\large\frac{5}{x+2}\big]^{-\large\frac{(x+2)}{5}}$
$\Rightarrow e^{\lim\limits_{x\to \infty}-\large\frac{5}{1+2/x}}$
$\Rightarrow e^{-5}$
Hence (c) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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