Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f\left(x\right)=\left\{\begin{array}{ll}\frac{{\log }_e\left(1-x+x^2\right)+{\log }_e\left(1+x+x^2\right)}{sec x-\cos x},& x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k& , x=0\end{array}\right.$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(\ln \left(1+x^2+x^4\right)\right)\cos x}{1-{\cos }^{2}x}=k$

Taking L.H.S$=\underset{x\to 0}{\lim }\frac{\left(\ln \left(1+x^2+x^4\right)\right)\cos x}{1-{\cos }^{2}x}$

$=\underset{x\to 0}{\lim }\frac{\left(\frac{\ln \left(1+x^2+x^4\right)}{x^2+x^4}\right)x^2\left(1+x^2\right)\cos x}{\left(\frac{{\sin }^{2}x}{x^2}\right)x^2}$

$=\underset{x\to 0}{\lim }\frac{\left(\frac{\ln \left(1+x^2+x^4\right)}{x^2+x^4}\right)\left(1+x^2\right)\cos x}{\left(\frac{{\sin }^{2}x}{x^2}\right)}$

$=\frac{\left(1\right)\left(1+0\right)1}{\left(1\right)}=1$

Now equating with R.H.S we get, $k=1$