$\underset{x→0}{\lim}\frac{\sqrt{1-x+x^2}-\sqrt{1+x^2}}{4^x-1}$ is equal to

$-\frac{1}{\log 16}$

$\underset{x→0}{\lim}\frac{\sqrt{1-x+x^2}-\sqrt{1+x^2}}{4^x-1}$

$\underset{x→0}{\lim}\frac{(\sqrt{1-x+x^2}-\sqrt{1+x^2})}{\sqrt{1-x+x^2}+\sqrt{1+x^2}}.\frac{(\sqrt{1-x+x^2}+\sqrt{1+x^2})}{\frac{4^x-1}{x}x}$

$\underset{x→0}{\lim}\frac{(1-x+x^2)-(1+x^2)}{2}.\frac{1}{\underset{x→0}{\lim}\frac{4^x-1}{x}x}=\frac{-x}{2\underset{x→0}{\lim}\frac{4^x-1}{x}x}=\frac{-1}{2\log 4}=-\frac{1}{\log 16}$