Question

$\underset{x\to 0}{lim}{(2^{x-1}+\frac{1}{2})}^{\frac{1}{x}}$ is equal to

• A. $\sqrt{2}$
• B. $\frac{1}{2}ln2$
• C. $ln2$
• D. $2$

Answer 1

The correct option is A $\sqrt{2}$

Let $A=\underset{x\to 0}{Lim}(2^{x-1}+\frac{1}{2})\left(1^{\infty }form\right)$
$=e^{\underset{x\to 0}{Lim}\frac{1}{x}[2^{x-1}+\frac{1}{2}-1]}=e^{\underset{x\to 0}{Lim}\left(\frac{2^x-1}{2x}\right)}=e^{\frac{1}{2}ln2}=\sqrt{2}$