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 $8$

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

It is of the form $\frac{0}{0}$, so applying L-Hospital's rule

$=\underset{x\to 2}{lim}\frac{2^x\log 2-2^{(3-x)}\log 2}{\left(\frac{-1}{2}\right)2^{\frac{-x}{2}}\log 2+2^{1-x}\log 2}$

$=\underset{x\to 2}{lim}\frac{2^x-2^{(3-x)}}{\left(\frac{-1}{2}\right)2^{\frac{-x}{2}}+2^{1-x}}$

$=\frac{4}{\frac{1}{2}}=8$