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$\lim\limits_{x\to \infty}\big[\large\frac{x^2+5x+3}{x^2+x+3}\big]^x$

$(a)\;e^4\qquad(b)\;e^2\qquad(c)\;e^3\qquad(d)\;1$

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$\lim\limits_{x\to \infty}\big[\large\frac{x^2+5x+3}{x^2+x+2}\big]^x=$$\lim\limits_{x\to \infty}\big[1+\large\frac{4x+1}{x^2+x+2}\big]^x$
$\Rightarrow \lim\limits_{x\to \infty}\big[\big(1+\large\frac{4x^2+x}{x^2+x+2}\big)^{\Large\frac{x^2+x+2}{4x+1}}\big]^{\Large\frac{(4x+1)x}{x^2+x+2}}$
$\lim\limits_{x\to \infty}(1+\lambda x)^{1/x}=e^{\lambda}$
$\Rightarrow e^{\large\lim\limits_{x\to \infty}\Large\frac{4x^2+x}{x^2+x+2}}$
$\Rightarrow e^{\large\lim\limits_{x\to \infty}\Large\frac{4+(1/x)}{1+(1/x)+(2/x^2)}}$
$\Rightarrow e^4$
Hence (a) is the correct answer.
answered Dec 31, 2013 by sreemathi.v
 

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