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$\lim\limits_{x\to 0}\bigg[\large\frac{1+5x^2}{1+3x^2}\bigg]^{\Large\frac{1}{x^2}}$

$(a)\;e^3\qquad(b)\;e^2\qquad(c)\;e^4\qquad(d)\;e^5$

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$\lim\limits_{x\to 0}[f(x)]^{g(x)}=e^{\large\lim\limits_{x\to 0}g(x)\log f(x)}$
$\lim\limits_{x\to 0}\big[\large\frac{1+5x^2}{1+3x^2}\big]^{\Large\frac{1}{x^2}}=e^{\lim\limits_{x\to 0}\large\frac{1}{x^2}\log\big[\Large\frac{1+5x^2}{1+3x^2}\big]}$
$\Rightarrow e^{\lim\limits_{\large x\to 0}\big[5.\Large\frac{\log(1+5x^2)}{5x^2}-\frac{3\log(1+3x^2)}{3x^2}\big]}$
$\Rightarrow e^{5-3}$
$\Rightarrow e^2$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 
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