Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Evaluate $\displaystyle\lim _{x \rightarrow 2}\left(\frac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}-2^{1-x}}}\right)$

Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow 2}\left(\frac{2^{x}+2^{3-x}-6}{\sqrt{2^{-x}}+2^{1-x}}\right)$
$=\displaystyle\lim _{x \rightarrow 2} \frac{2^{2 x}+2^{3}-6\left(2^{x}\right)}{\sqrt{2^{x}}-2}$
$=\displaystyle\lim _{x \rightarrow 2} \frac{\left(2^{x}-2\right)\left(2^{x}-4\right)\left(\sqrt{2^{x}}+2\right)}{2^{x}-4}$
$=\displaystyle\lim _{x \rightarrow 2}\left[\left(2^{x}-2\right)\left(\sqrt{2^{x}}+2\right)\right]$
$=8$