Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $\infty$

Answer 1

The correct option is A $1$

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

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

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

$=1.(\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1)$