Question

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

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

The correct option is C $8$

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

$\underset{x\to 2}{lim}\frac{2^x+\frac{8}{2^x}-6}{\sqrt{\frac{1}{2x}}-\frac{-2}{2^x}}$

$=\underset{x\to 2}{lim}\frac{\frac{1}{2x}\left[\right(2^x{)}^2-{6.2}^x+8]}{\frac{1}{2^x}[\sqrt{2^x}-2]}$

$=\underset{x\to 2}{lim}\frac{(2^x{)}^2-{6.2}^x+8}{\sqrt{2^x}-2}$

put $2^x=t$

As $x+2,t\to 4$

$=\underset{t\to 4}{lim}\frac{(t^2-6t+8)\times (\sqrt{t}+2)}{(\sqrt{t}-2)(\sqrt{t}+2)}$

$=\underset{t\to 4}{lim}\frac{(t-4)(t-2)}{(t-4)}(\sqrt{t}+2)$

$=2\times (2+2)$

$=2\times 4$

$=8$