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$\lim\limits_{x\to a}\bigg($ $\bigg[$ $\big(\large\frac{a^{1/2}+x^{1/2}}{a^{1/4}-x^{1/4}}\big)^{-1}$ $-$ $\frac{2(ax)^{1/4}}{x^{3/4}-a^{1/4}x^{1/2}+a^{1/2}x^{1/4}-a^{3/4}}\bigg]^{-1}$ $-$ $(\sqrt 2)^{\large\log_4 a}\bigg)^8$ is

$(a)\;a\qquad(b)\;a^{3/4}\qquad(c)\;a^2\qquad(d)\;a^3$

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1 Answer

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Put $b=a^{1/4}$
$y=x^{1/4}$
The given limit can be written as
$\lim\limits_{y=a^{1/4}}\bigg(\bigg[\big(\large\frac{b^2+y^2}{b-y}\big)^{-1}-\frac{2by}{y^3-by^2+b^2y-b^3}\bigg]^{-1}-$$2^{\large 1/2\log_4 b^4}\bigg)^8$
$\Rightarrow \lim\limits_{y\to a^{1/4}}\bigg(\bigg[\large\frac{b-y}{b^2+y^2}-\frac{2by}{(y-b)(b^2+y^2}\bigg]^{-1}-$$b\bigg)^8$
$\Rightarrow \lim\limits_{\large y\to a^{1/4}}\bigg(\big[\large\frac{1}{b-y}\big]^{-1}$$-b\bigg)^8$
$\Rightarrow \lim\limits_{\large y\to a^{1/4}}y^8=a^2$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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