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Q. $\displaystyle\lim _{x \rightarrow 0} \frac{\log \left(1+x+x^{2}\right)+\log \left(1-x+x^{2}\right)}{\sec x-\cos x}$ is equal to

Limits and Derivatives

Solution:

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