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$\lim\limits_{x\to 0}\big(\large\frac{x^2+5x+3}{x^2+x+2}\big)^x$ =

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

1 Answer

$\lim\limits_{x\to \infty}\big(\large\frac{x^2+5x+3}{x^2+x+2}\big)^x$
$\Rightarrow \lim\limits_{x\to \infty}\big(1+\large\frac{4x+1}{x^2+x+2}\big)^x$
$\lim\limits_{x\to \infty}\bigg[\big(1+\large\frac{4x+1}{x^2+x+2}\big)^{\Large\frac{x^2+x+2}{4x+1}}\bigg]^{\Large\frac{x(4x+1)}{x^2+x+1}}$
$\Rightarrow \lim\limits_{\large e^{x\to \infty}}\large\frac{x(4x+1)}{x^2+x+2}$
$\Rightarrow e^4$
Hence (a) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 
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