Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x}=\underset{x\to 0}{lim}\frac{\log [{(1+x^2)}^2-x^2]}{\frac{(1-{\cos }^{2}x)}{\cos x}}=\underset{x\to 0}{lim}\frac{\log (1+x^2+x^4)}{\frac{{\sin }^{2}x}{\cos x}}=\underset{x\to 0}{lim}\frac{\log (1+x^2(1+x^2))}{x^2(1+x^2)}\cdot x^2(1+x^2)\cdot \frac{\cos x}{{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{\log (1+x^2(1+x^2))}{x^2(1+x^2)}\cdot \left(\cos x\right)\cdot (1+x^2)\cdot \frac{x^2}{{\sin }^{2}x}(\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1,\underset{x\to 0}{lim}\frac{\sin x}{x}=1)=\underset{x\to 0}{lim}\left(\cos x\right)\cdot (1+x^2)=1$