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

Answer: (B)

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