Question

The value of $\underset{x\to 2}{lim}\frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}$ is

• A. $16$
• B. $8$
• C. $4$
• D. $2$

Answer 1

The correct option is B $8$

$\underset{x\to 2}{lim}\frac{2^x+2^{3-x}-6}{\sqrt{2^{-x}}-2^{1-x}}$
$=\underset{x\to 2}{lim}\frac{(2^x{)}^2-6\times 2^x+2^3}{\sqrt{2^x}-2}$ [Multiplying $N^r$and $D^r$ by $2^x$]
$=\underset{x\to 2}{lim}\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{(\sqrt{2^x}-2)(\sqrt{2^x}+2)}$
$=\underset{x\to 2}{lim}\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{(2^x-4)}$
$=\underset{x\to 2}{lim}(2^x-2)(\sqrt{2^x}+2)=(2^2-2)(2+2)=8$