Question

$\underset{x\to a}{lim}{\left\{[{\left(\frac{a^{1/2}+x^{1/2}}{a^{1/4}-x^{1/4}}\right)}^{-1}-\frac{2(ax{)}^{1/4}}{x^{3/4}{a}^{1/4}{x}^{1/2}+a^{1/2}{x}^{1/4}}-a^{3/4}-\sqrt{2^{lo{g}_{4}a}}]\right\}}^8$is

• A. a
• B. $a^{3/4}$
• C. $a^2$
• D. none of these

Answer 1

Answer: (C)

$a^2$

Simplifying the expression in brackets by setting $a^{1/4}=b$ and $x^{1/4}=y,$ the function whose limit is required can be written as
${\{{[{\left(\frac{b^2+y^2}{b-y}\right)}^{-1}-\frac{2by}{y^3-b{y}^2+b^{2}y-b^3}]}^{-1}-2^{\frac{1}{2}lo{g}_4{b}^4}\}}^8$
$={\{{[\frac{b-y}{b^2+y^2}-\frac{2by}{(y-b)(b^2+y^2)}]}^{-1}-b\}}^8$
$={\left\{[\frac{1}{b-y}-b]\right\}}^8=y^8=x^2$
Hence the required lim as $x\to a$ is $a^2$