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)

8

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

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

$=\underset{x\to 2}{lim}\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{(\sqrt{2^x}-2)(\sqrt{2^x}+2)}$
$=\underset{x\to 2}{lim}\frac{(2^x-4)(2^x-2)(\sqrt{2^x}+2)}{2^x-4}$
$=\underset{x\to 2}{lim}(2^x-2)(\sqrt{2^x}+2)=(2^2-2)(2+2)=8$
Hence, option 'B' is correct.