Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is B $1$

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

$=\underset{x\to 0}{lim}\frac{\log \left[\right(1+x^2{)}^2-x^2]}{(1-{\cos }^{2}x)/\cos x}$

$=\underset{x\to 0}{lim}\frac{\log (1+x^2+x^4)}{\sin x\tan x}$

$=\underset{x\to 0}{lim}\frac{\log (1+x^2(1+x^2\left)\right)}{x^2(1+x^2)}.x^2(1+x^2).\frac{1}{\frac{\sin x}{x}.\frac{\tan x}{x}.x^2}=1$

Since $\left(\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1\right)$