Question

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

• A. 1
• B. -1
• C. 0
• D. $\infty$

Answer 1

Answer: (A)

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