$\underset{x→0}{\lim}\frac{\log(1+x+x^2)+\log(1-x+x^2)}{\sec x-\cos x}$ equal to

1

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

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

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